Obstacle Problems Involving The Fractional Laplacian

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1 Obstacle Problems Involving The Fractional Laplacian Donatella Danielli and Sandro Salsa January 27, Introduction Obstacle problems involving a fractional power of the Laplace operator appear in many contexts, such as in pricing of American options governed by assets evolving according to jump processes [25], or in the study of local minimizers of some nonlocal energies [24]. In the first part of this expository paper we are concerned with the stationary case, that can be stated in several ways. Given a smooth function ϕ : R n R, n > 1, with bounded support (or at least rapidly decaying at infinity), we look for a continuous function u satisfying the following system: u ϕ in R n ( ) s u 0 in R n ( ) s (1) u = 0 when u > ϕ u (x) 0 as x +. Here we consider only the case s (0, 1). The set Λ (u) = {u = ϕ} is called the contact or coincidence set. The free boundary is the set F (u) = Λ (u). The main theoretical issues in a constrained minimization problem are optimal regularity of the solution and the analysis of the free boundary. If s = 1 and R n is replaced by a bounded domain Ω our problem corresponds to the usual obstacle problem for the Laplace operator. The existence of a unique solution satisfying some given boundary condition u = g can be obtained by minimizing the Dirichlet integral in H 1 (Ω) under the constraint u ϕ. The solution is the least superharmonic function greater or equal to ϕ in Ω, with u g on Ω, and inherits up to a certain level the regularity of ϕ ([32]). In fact, even if ϕ is smooth, u is only C 1,1 loc, which is the optimal regularity. A classical D.D. was supported in part by NSF grant DMS

2 reference for the obstacle problem, including the regularity and the complete analysis of the free boundary is [18]. See also the recent book [55]. Analogously, the existence of a solution u for problem (1) can be obtained by variational methods as the unique minimizer of the functional J (v) = R n R n 2 v (x) v (y) x y n+2s dxdy over a suitable set of functions v ϕ. We can also obtain u via a Perron type method, as the least supersolution of ( ) s such that u ϕ. By analogy (see also later the Signorini problem), when ϕ is smooth, we expect the optimal regularity for u to be C 1,s. This is indeed true, as it is shown by Silvestre in [62] when the contact set {u = ϕ} is convex and by Caffarelli, Salsa, Silvestre in [22] in the general case. The case s = 1/2 is strongly related to the so called thin (or lower dimensional) obstacle problem for the Laplace operator. To keep a connection with the obstacle problem for ( ) s, it is better to work in R n+1, writing X = (x, y) R n R. The thin obstacle problem concerns the case in which the obstacle is not anymore n + 1 dimensional, but supported instead on a smooth n dimensional manifold M in R n+1. This problem and variations of it also arise in many applied contexts, such as flow through semi-permeable membranes, elasticity (known as the Signorini problem, see below), boundary control temperature or heat problems (see [28]). More precisely, let Ω be a domain in R n+1 divided into two parts Ω + and Ω by M. Let ϕ : M R be the (thin) obstacle and g be a given function on Ω satisfying g > ϕ on M Ω. The problem consists in the minimization of the Dirichlet integral J (v) = v 2 over the closed convex set K = { v H 1 (Ω) : v = g on Ω and v ϕ on M Ω }. Since we can perturb the solution u upwards and freely away from M, it is apparent that u is superharmonic in Ω and harmonic in Ω\M. One expects the continuity of the first derivatives along the directions tangential to M, and the one sided continuity of normal derivatives ([32]). In fact (see [16]), on M, u satisfies the following complementary conditions u ϕ, u ν + + u ν 0, (u ϕ) (u ν + + u ν ) = 0 where ν ± are the interior unit normals to M from the Ω ± side. The free boundary here is given by the boundary of the set Ω\Λ (u) in the relative topology of M, and in general, we expect it is a (n 1) dimensional manifold. Ω 2

3 As mentioned above, a related problem is the Signorini problem 1 (or boundary thin obstacle problem), in which the manifold M is part of Ω and one has to minimize the Dirichlet integral over the closed convex set K = { v H 1 (Ω) : v = g on Ω\M and v ϕ on M }. In this case, u is harmonic in Ω and on M it satisfies the complementary conditions u ϕ, u ν + 0, (u ϕ) u ν + = 0. If M is a hyperplane (say {y = 0}) and Ω is symmetric with respect to M, then the thin obstacle in Ω and the boundary obstacle problems in Ω + or Ω are equivalent. Let us see how these problems are related to the obstacle problem for the ( ) 1/2. This is explained through the following remarks. (a) Reduction to a global problem. Let Ω = B 1 be the unit ball in R n+1 and B 1 = B 1 {y = 0}. Let ϕ : R n R be a smooth obstacle, ϕ < 0 on B 1 and positive somewhere inside B 1. Consider the following Signorini problem in B 1 + = B 1 {y > 0}: u = 0 in B 1 + u = 0 on B 1 {y > 0} u (x, 0) ϕ (x) in B 1 u y (x, 0) 0 in B (2) 1 u y (x, 0) = 0 when u (x, 0) > ϕ (x). We want to convert the above problem in B 1 into a global one, that is in R n (0, + ). To do this, let η be a radially symmetric cut-off function in B 1 such that {ϕ > 0} {η = 1} and supp (η) B 1. Extending ηu by zero outside B 1, we have η (x) u (x, 0) ϕ (x) and also (ηu) y (x, 0) 0 for every x R n. Moreover, (ηu) y (x, 0) = 0 if η (x) u (x, 0) > ϕ (x). Let now v be the unique solution of the following Neumann problem in the upper half space, vanishing at infinity: { v = (ηu) in R n {y > 0} v y (x, 0) = 0 in R n. Then w = ηu v is a solution of a global Signorini problem with ϕ v as the obstacle. Thus, the regularity of u in the local setting can be inferred from the regularity for the global problem. The opposite statement is obvious. (b) Realization of ( ) 1/2 as a Dirichlet-Neumann map. Consider a smooth function u 0 : R n R with rapid decay at infinity. Let u : R n (0, + ) R be the unique solution of the Dirichlet problem { u = 0 in R n (0, + ) u (x, 0) = u 0 (x) in R n 1 After Fichera, see ([30]),

4 vanishing at infinity. We call u the harmonic extension of u 0 to the upper half space. Consider the Dirichlet-Neumann map T : u 0 (x) u y (x, 0). We have: (T u 0, u 0 ) = u y (x, 0) u (x, 0) dx R n { = u (X) u (X) + u (X) 2} dx = R n (0,+ ) R n (0,+ ) u (X) 2 dx 0 so that T is a positive operator. Moreover, since u 0 is smooth and u y is harmonic, we can write: We conclude that As a consequence: T T u 0 = y ( y )u (x, 0) = u yy (x, 0) = u 0. T = ( ) 1/2. 1. If u = u (X) is a solution of the Signorini problem in R n (0, + ), that is u = 0 in R n+1, u (x, 0) ϕ, u y (x, 0) 0, and (u ϕ) u y (x, 0) = 0 in R n, then u 0 = u (, 0) solves the obstacle problem for ( ) 1/2. 2. If u 0 is a solution of the obstacle problem for ( ) 1/2, then its harmonic extension to R n (0, + ) solves the corresponding Signorini problem. Therefore, the two problems are equivalent and any regularity result for one of them can be carried to the other one. More precisely, consider the optimal regularity for the solution u 0 of the obstacle problem for ( ) 1/2, which is C 1,1/2. If we can prove a C 1,1/2 regularity of the solution u of the Signorini problem up to y = 0, then the same is true for u 0. On the other hand, the C 1,α regularity of u 0 extends to u, via boundary estimates for the Neumann problem. Similarly, the analysis of the free boundary in the Signorini problem carries to the obstacle problem for ( ) 1/2 as well and vice versa. Although the two problem are equivalent, there is a clear advantage in favor of the Signorini type formulation. This is due to the possibility of avoiding the direct use of the non local pseudodifferential operator ( ) 1/2, by localizing the problem and using local PDE methods, such as monotonicity formulas and classification of blow-up profiles. At this point it is a natural question to ask whether there exists a PDE realization of ( ) s for every s (0, 1), s 1 2. The answer is positive as it is shown by Caffarelli and Silvestre in [23]. Indeed in a weak sense we have that ( ) s u 0 (x) = κ a lim y 0+ ya u y (x, y) 4

5 for a suitable constant κ a, where u is the solution of the problem { La u = div (y a u) = 0 in R n (0, + ) u (x, 0) = u 0 (x) in R n vanishing at infinity, where a = 2s 1. Coherently, we call u the L a harmonic extension of u 0. Thus problem (1) is equivalent to the following Signorini problem for the operator L a = div(y a ), u (x, 0) ϕ in R n (3) L a u = div (y a u) = 0 in R n (0, + ) (4) lim y 0+ y a u y (x, y) = 0 when u (x, 0) > ϕ (x) (5) lim y 0+ y a u y (x, y) 0 in R n. (6) For y > 0, u is smooth so that (4) is understood in the classical sense. The equations at the boundary (5) and (6) should be understood in a weak sense. Since in [62] it is shown that u (x, 0) C 1,α for every α < s, for the range of values 2s 1 < α < s, lim y 0+ y a u y (x, y) can be understood in the classical sense too. The solution u of the above Signorini problem can be extended to the whole space by symmetrization, setting u (x, y) = u (x, y). Then, by the results in [23], condition (5) holds if and only if the extended u is a solution of L a u = 0 across y = 0, where u (x, 0) > ϕ (x). On the other hand, condition (6) is equivalent to L a u 0 in the sense of distributions. Thus, for the extended u, the Signorini problem translates into the following system: u (x, 0) ϕ (x) in R n u (x, y) = u (x, y) in R n+1 L a u = 0 in R n+1 \ {(x, 0) : u (x, 0) = ϕ (x)} L a u 0 in R n+1, in the sense of distributions. with u vanishing at infinity. Again, we can exploit the advantages to analyze the obstacle problem for a nonlocal operator in PDE form by considering a local version of it. Indeed, to study the optimal regularity properties of the solution we will focus on the following local version, where ϕ : B R: u (x, 0) ϕ (x) in B 1 u (x, y) = u (x, y) in B 1 L a u = 0 in B 1 \ {(x, 0) : u (x, 0) = ϕ (x)} L a u 0 in B 1, in the sense of distributions. The above problem can be thought of as the minimization of the weighted Dirichlet integral J a (v) = y a u (X) 2 dx B 1 5

6 over the set K a = { v W 1,2 (B 1, y a ) : u (x, 0) ϕ (x) }. In a certain sense, this corresponds to an obstacle problem, where the obstacle is defined in a set of codimension 1 + a, where a is not necessarily an integer. The operator L a is degenerate elliptic, with a degeneracy given by the weight y a (. This weight belongs to the Muckenhoupt class A ) 2 R n+1 (. We recall that a positive weight function w = w (X) belongs to A ) 2 R N if ( ) ( ) 1 1 w w 1 C B B B B for every ball B R N. For the class of degenerate elliptic operators of the form Lu =div(a (X) u), where λw (X) ξ 2 A (X) ξ ξ Λw (X) ξ 2, there is a well established potential theory for solutions in the weighted Sobolev space W 1,2 (Ω, w) (Ω bounded domain in R N ), defined as the closure of C ( Ω) in the norm [ v 2 w + Ω Ω ] 1/2 v 2 w (see [29]). Since w A 2, the gradient of a function in W 1,2 (Ω, w) is well defined in the sense of distributions and belongs to the weighted space L 2 (Ω, w). The outline of the first part. We intend here to give a brief review of the results concerning the analysis of the solution and the free boundary of the obstacle for the fractional Laplacian, mainly based on the extension method. For the thin obstacle problem, Caffarelli in [16] proves that u is C 1,α up to y = 0, for some α 1/2. Subsequently Athanasopoulos and Caffarelli achieve the optimal regularity of the solution in [7]. In the case of zero obstacle, the regularity of the free boundary around a so called nondegenerate (or stable or regular) point is analyzed by Athanasopoulos, Caffarelli and Salsa in [10]. Indeed these last two papers opened the door to all subsequent developments. In [36], Garofalo and Petrosyan start the analysis of F (u) around non regular points (also for non zero obstacles). They obtain a stratification result for singular points, i.e. points of F (u) of vanishing density for Λ (u), in terms of homogeneity of suitable blow-ups of the solution. The analysis of the obstacle problem for the operator ( ) s, 0 < s < 1, starts with Silvestre in ([62]), which shows C 1,α estimates for the solution for any α < s and also α = s if the interior of the coincidence set is convex. Notably, Silvestre does not use any extension properties; his methods are purely nonlocal. A few years later, Caffarelli, Salsa and Silvestre ([22]) extend to the fractional Laplacian case the results in [10] on the optimal regularity and the analysis of the regular part of the free boundary. Recently, in [12], Barrios, Figalli and Ros-Oton continued the work of [36], giving a complete picture of the free boundary under two basic assumptions. 6

7 The first one is a strict concavity of the obstacle, the same assumption needed in the case of the classical obstacle problem. The second one prescribes zero boundary values of the solution and it turns out to be crucial. Here, we shall focus mainly on the optimal regularity of the solution and on the analysis and structure of the free boundary, only mentioning, for brevity reasons, recent important results on higher regularity and extension to more general operators. In particular, we will give here an outline of the strategy used in the papers ([22]) and [12]. A few comments on the key concepts and tools that will repeatedly appear are in order. Semi-convexity: it is a peculiarity of solutions of the obstacle problem. More precisely, semi-convexity along tangential directions τ (i.e. parallel to the plane y = 0) and semiconcavity along the y direction consistently play a key role. It is noteworthy that, for global solution of the zero thin obstacle case, the tangential semi-convexity of u comes for free, since u (X + hτ) and u (X hτ) are admissible nonnegative superharmonic functions, and therefore 1 (u (X + hτ) + u (X hτ)) u (X). 2 Asymptotic profiles. From semi-convexity, one deduces that suitable global asymptotic profiles coming from blow-ups of u around a free boundary point (say, the origin) are tangentially convex and can be classified according to their homogeneity degree. From this it is an easy matter to deduce optimal regularity. Frequency and monotonicity formulas. Frequency formulas of Almgren type, first introduced in the case s = 1/2 in ([10]) are key tools in carrying optimal regularity from global to local solutions. Other types of monotonicity formulas, such as Weiss or Monneau-types, first introduced in ([36]) for s = 1/2, play a crucial role in the analysis of non-regular points of the free boundary. Carleson estimates and boundary Harnack principles are by now standard tools in the study of the optimal regularity of the free boundary, in our case around the so called regular points. Due to the non homogeneous right hand side in the equation, the Carleson estimate and boundary Harnack principle proved here are somewhat weaker than the usual ones. More recently, De Silva and Savin ([27]) have applied these principles to prove higher regularity of the free boundary. We will always assume that the origin belongs to the free boundary. The outline of the second part. In Section 3 we consider two timedependent models, which can be thought of as parabolic counterparts of the systems (1) and (2). In the first part, Section 3.1, we discuss the regularity of solutions to the parabolic fractional obstacle problem { min{ v t + ( ) s v, v ψ} = 0 on [0, T ] R n v(t ) = ψ on R n, following [19]. In particular, under some assumptions on the obstacle ψ, the solution v is shown to be globally Lipschitz continuous in space-time. Moreover, 7

8 v t and ( ) s v belong to suitable Hölder and loglipschitz spaces. The regularity in space is optimal, whereas the regularity in time is almost optimal in the cases s = 1/2 and s 1. In Section 3.2 we give an overview of the parabolic Signorini problem v t v = 0 in Ω T := Ω (0, T ], v φ, ν v 0, (v φ) ν v = 0 on M T := M (0, T ], v = g on S T := S (0, T ], v(, 0) = φ 0 on Ω 0 := Ω {0}. Here S = Ω \ M. Similarly to the elliptic case, we are interested in the regularity properties of v, and the structure and regularity of the free boundary Γ(v) = MT {(x, t) M T v(x, t) > φ(x, t)}, where MT indicates the boundary in the relative topology of M T. Following [26], the analysis comprises the monotonicity of a generalized frequency function, the study of blow-ups and the ensuing regularity of solutions, the classification of free boundary points, and the regularity of the free boundary at so-called regular points. 2 The obstacle problem for the fractional Laplacian This section is devoted to the study of the fractional Laplacian obstacle problem that we recall below. Given a smooth function ϕ : R n R, with bounded support (or rapidly vanishing at infinity), we look for a continuous function u satisfying the following conditions: u ϕ in R n ( ) s u 0 in R n ( ) s u = 0 when u > ϕ u (x) 0 as x +. We list below the main steps in the analysis of the problem that we are going to describe. 1. Construction of the solution and basic properties. 2. Lipschitz continuity, semiconvexity and C 1,α estimates. 3. Reduction to the thin obstacle for the operator L a. 4. Optimal regularity for tangentially convex global solution. 5. Classification of asymptotic blow-up profiles around a free boundary point. 8

9 6. Optimal regularity of the solution. 7. Analysis of the free boundary at stable points: Lipschitz continuity. 8. Boundary Harnack Principles and C 1,α regularity of the free boundary at stable points. 9. Structure of the free boundary. Steps 1, 2 and 3 are covered in [62]. In particular from [62] we borrow the C 1,α estimates without proof, for brevity. The steps 3-8 follow basically the paper [22], except for step 2 and part of 5. In particular, the optimal regularity of global solutions follows a different approach, similar to the corresponding proof for the zero obstacle problem in [10]. Step 8 is taken from [12]. Finally, step 9 comes from [10] (Carleson estimate) and [22] (Boundary Harnack). 2.1 Construction of the solution and basic properties We start by proving the existence of a solution. Observe that the proof fails for n = 1 and s > 1/2, because in this case it is impossible to have ( ) s u 0 in R with u vanishing at infinity. Let S be the set of rapidly decreasing C functions in R n. We denote by Ḣ s the completion of S in the norm f 2 Ḣ = s R n R n 2 f (x) f (y) x y n+2s dxdy ξ 2s ˆf (ξ) 2 dξ. R n Equipped with the inner product (f (x) f (y)) (g (x) g (y)) (f, g)ḣs = R n R n x y n+2s dxdy = 2 f (x) ( ) s g (x) dx ξ 2s ˆf (ξ) ĝ (ξ)dξ, R n R n Ḣ s is a Hilbert space. Since we are considering n 2 and s < 1 n/2, it turns out that Ḣs coincides with the set of functions in L 2n/(n 2s), for which the Ḣs norm is finite. The solution u 0 of the obstacle problem is constructed as the unique minimizer of the strictly convex functional J (v) = v 2 Ḣ s over the closed, convex set K s = { v Ḣs : v ϕ }. In the following proposition we gather some basic properties of u. Proposition 1 Let u 0 be the minimizer of the functional J over K s. Then: 9

10 (a) The function u 0 is a supersolution, that is ( ) s u 0 0 in R n in the sense of measures. Thus, it is lower semicontinuous and, in particular, the set {u 0 > ϕ} is open. (b) u 0 is actually continuous in R n. (c) If u 0 (x) > ϕ (x) in some ball B then ( ) s u 0 = 0 in B. Proof. (a) Let h 0 be any smooth function with compact support. If t > 0, u 0 + th ϕ so that or from which (u 0, u 0 )Ḣs (u 0 + th, u 0 + th)ḣs 0 2t (u 0, h)ḣs + t2 (h, h)ḣs = (( ) s h) L 2 + t 2 (h, h)ḣs (u 0, ( ) s h) L 2 = (( ) s u 0, h) L 2 0. Therefore ( ) s u 0 is a nonnegative measure and therefore is lower semicontinuous by Proposition A1. (b) The continuity follows from Proposition A2. (c) For any test function h 0, supported in B the proof in (a) holds also for t < 0. Therefore ( ) s u 0 = 0 in B. Corollary 2 The minimizer u 0 of the functional J over K s is a solution of the obstacle problem. 2.2 Lipschitz continuity and semiconvexity and C 1,α estimates Following our strategy, we first show that, if ϕ is smooth enough, then the solution of our obstacle problem is Lipschitz continuous and semiconvex. We are mostly interested in the case ϕ C 1,1. When ϕ has weaker regularity, u 0 inherits corresponding weaker regularity (see [62]). We emphasize that the proof in this subsection depends only on the maximum principle and translation invariance. Lemma 3 The function u 0 is the least supersolution of ( ) s such that u 0 > ϕ and lim inf x u 0 (x) 0. Proof. Let v such that ( ) s v 0, v > ϕ and lim inf x v (x) 0. Let w = min {u 0, v}. Then w is lower-semicontinuous in R n and is another supersolution above ϕ (by Propositions A1 and A4). We show that w u 0. Since ϕ w u 0, we have w (x) = u 0 (x) for every x in the contact set Λ (u 0 ) = {u 0 = ϕ}. In Ω = {u 0 > ϕ}, u 0 solves ( ) s u 0 = 0 and w is a supersolution. By Proposition 1 (b), u 0 is continuous. Then w u 0 is lowersemicontinuous and w u 0 from comparison. 10

11 Corollary 4 The function u 0 is bounded and sup u 0 sup ϕ. If the obstacle ϕ has a modulus of continuity ω, then u 0 has the same modulus of continuity. In particular, if ϕ is Lipschitz, then u 0 is Lipschitz and Lip(u 0 ) Lip(ϕ). Proof. By hypothesis u 0 ϕ. The constant function v (x) = sup ϕ is a supersolution that is above ϕ. By Lemma 3, u 0 v in R n. Moreover, since ω is a modulus of continuity for ϕ, for any h R n, ϕ (x + h) + ω ( h ) ϕ (x) for all x R n. Then, the function u 0 (x + h) + ω ( h ) is a supersolution above ϕ (x). Thus u 0 (x + h) + ω ( h ) u 0 (x) for all x, h R n. Therefore u 0 has a modulus of continuity not larger than ω. Lemma 5 Let ϕ C 1,1 and assume that inf ττ ϕ C, for any unit vector τ. Then ττ u 0 C too. In particular, u 0 is semiconvex. Proof. Since ττ ϕ C, we have ϕ (x + hτ) + ϕ (x hτ) 2 for every x R n and h > 0. Therefore: V (x) u 0 (x + hτ) + u 0 (x hτ) 2 + Ch 2 ϕ (x) + Ch 2 ϕ (x) and V is also a supersolution: ( ) s V 0. Thus, by Lemma 3, V u so that: u 0 (x + hτ) + u 0 (x hτ) 2 + Ch 2 u 0 (x) for every x R n and h > 0. This implies ττ u 0 C. From the results in [62] we can prove a partial regularity result, under the hypothesis that ϕ is smooth. Theorem 6 Let ϕ C 2. Then u 0 C 1,α for every α < s and ( ) s u 0 C β for every β < 1 s. The proof is long and very technical, so we refer to the original paper [62]. However, an idea of the proof in the case of the zero thin obstacle problem can be given without much effort. Indeed, from tangential semi-convexity we deduce (here u is harmonic outside Λ (u)) u yy C in B 1 \Λ (u). In particular, the function u y Cy is monotone and bounded. Thus we are allowed to define in B 1 : σ (x) = lim y 0+ u y (x, y). Since, in this case 0 u = 2u y H n Λ(u) 11

12 in the sense of measures, we have σ (x) 0 in Λ (u) and by symmetry, σ (x) = 0 in B 1\Λ (u). We may summarize the properties of the solution of a zero thin obstacle problem in complementary form as follows: u 0, u u = 0 in B 1 u = 0 in B 1 \Λ (u) u (x, 0) 0, σ (x) 0, u (x, 0) σ (x) = 0 in B 1 σ (x) = 0 in B 1\Λ (u) Now, let u be a solution of the zero thin obstacle problem in B 1, normalized by u L 2 (B = 1. To prove local 1) C1,α estimates, it is enough to show that σ C 0,α near the free boundary F (u). In fact, in the interior of Λ (u), u (x, 0) is smooth and so is σ. On the other hand, on Ω = B \Λ (u), σ = 0. Thus, if we show that σ is C 0,α in a neighborhood of F (u), then u C 1,α from both sides of the free boundary by standard estimates for the Neumann problem. In particular it is enough to show uniform estimates around a free boundary point, say the origin. This can be achieved by a typical iteration procedure, in the De Giorgi style. We distinguish two steps: Step 1: To show that near the free boundary we can locate large regions where σ grows at most linearly (estimates in measure of the oscillation of σ). Step 2: Using Poisson representation formula and semi-concavity, we convert the estimate in average of the oscillation of σ, done in step 1, into pointwise estimates, suitable for a dyadic iteration of the type u y (X) β k in B γ k (x 0) [ 0, γ k] (x 0 F (u) ) for some 0 < γ < 1, 0 < β < 1, and any k 0. The details can be found in the paper of Caffarelli [16] (see also the review paper [60]). 2.3 Thin obstacle for the operator L a. Local C 1,α estimates To achieve optimal regularity we now switch to the equivalent thin obstacle problem for the operator L a as mentioned in the introduction and that we restate here: u (x, 0) ϕ (x) in B 1 u (x, y) = u (x, y) in B 1 (7) L a u = 0 in B 1 \Λ (u) L a u 0 in B 1, in the sense of distributions. In the global setting (i.e. with B 1 replaced by R n ), u 0 (x) = u (x, 0) is the solution of the global obstacle problem for ( ) s and ( ) s u 0 = κ a lim y 0+ ya u y (x, y). 12

13 The estimates in Corollary 4 and Lemma 5 translate, after an appropriate localization argument and the use of boundary estimates for the operator L a, into corresponding estimates for the solution of u. Namely: Lemma 7 Let ϕ C 2,1 (B 1) and u be the solution of (7). Then 1. x u (X) C α ( B 1/2 ) for every α < s; 2. y a u y (X) C α ( B 1/2 ) for every α < 1 s; 3. u ττ (X) C in B 1/2. Proof. From Corollary 4 and Lemma 5 we have that the above estimates holds on y = 0. Since xj u and u ττ also solve the equation L a w = 0 in B 1 \Λ (u), the estimates 1 and 3 extend to the interior. On the other hand w (x, y) = y a u y (X) solves the conjugate equation div( y a w (X)) and we obtain 2. Remark 1 Observe that u can only be C 1,α in both variables up to y = 0 only if a 0. If a > 0, since y a u y (X) has a non-zero limit for some x in the contact set, it follows that u y cannot be bounded. We close this subsection with a compactness result, useful in dealing with blow-up sequences. Lemma 8 Let {v j } be a bounded sequence of functions in W 1,2 (B 1, y a ). Assume that there exists a constant C such that, in B 1 : x v j (X) C and y v j (X) C y a (8) and that, for each small δ > 0, v j is uniformly C 1,α in B 1 δ { y > δ}. Then, there exists a subsequence {v jk } strongly convergent in W 1,2 ( B 1/2, y a). Proof. From the results in [42], there is a subsequence, that we still call {v j }, that converges strongly in L 2 ( B 1/2, y a). Since for each δ > 0, v j is uniformly bounded in C 1,α in the set B 1 δ { y > δ}, we can extract a subsequence so that v j converges uniformly in B 1 δ { y > δ}. Thus, v j converges pointwise in B 1 \ {y = 0}. Now, from (8) and the fact that C and y a both belong to L 2 ( B 1/2, y a), the convergence of each partial derivative of v j in L 2 ( B 1/2, y a) follows from the dominated convergence theorem. 2.4 Minimizers of the weighted Rayleigh quotient and a monotonicity formula The next step towards optimal regularity is to consider tangentially convex global solutions. 13

14 Lemma 9 Let θ denote the surface gradient on the unit sphere B 1. Set, for 1 < a < 1, B λ 0,a = inf + θ w 2 y a ds 1 : w W 1,2 ( B + w 2 y a 1 ds, ya ds ) : w = 0 on ( B 1) + B + 1 where ( B 1) + = {(x, x n ) B 1, x n > 0}. Then the first eigenfunction, up to a multiplicative factor, is given by and 2 w (x, y) = ( x 2 n + y 2 x n ) s s = (1 a) /2 λ 0,a = 1 a 4 (2n + a 1). The following lemma gives a first monotonicity result. Lemma 10 Let w be continuous in B 1, w (0) = 0, w (x, 0) 0, w (x, 0) = 0 on Λ {y = 0}, L a w = 0 in B 1 \Λ. Assume that the set {x B r : w (x, 0) < 0} is non empty and convex. Set β (r) = β (r; w) = 1 r 1 a B + r y a w (X) 2 dx. X n+a 1 Then, β (r) is bounded and increasing for r (0, 1/2]. Proof. We have L a w 2 = 2wL a w + 2y a w 2 = 2y a w 2, so that β (r) = 1 y a w (X) 2 r 1 a B r + X n+a 1 dx = 1 ( L ) a w 2 2r 1 a B r + X Now: β (r) = a 1 2r 2 a B r + ( L ) a w 2 1 n+a 1 dx + X r n B + r n+a 1 dx. y a w 2 ds. Since w (0, 0) = 0 and y a w y (x, y) w (x, y) 0 as y 0 +, after simple computations, we obtain: β (2n + a 1) (r) (1 a) 4r n+2 y a w 2 ds + 1 B r + r n y a θ w 2 ds. B r + The convexity of {x B r : w (x, 0) < 0} implies that the Rayleigh quotient must be greater than λ 0,a and therefore we conclude β (r) 0 and, in particular, β (r) β (1/2). 2 Formally, the first eigenvalue can be obtained plugging α = s = (1 a) /2 and n + a instead of n into the formula α (α 1) + nα. 14

15 2.5 Optimal regularity for tangentially convex global solutions In this section we consider global solutions that represent possible asymptotic profiles, obtained by a suitable blow-up of the solution at a free boundary point. First of all we consider functions u : R n R R, homogeneous of degree k, solutions of the following problem: u (x, 0) 0 in R n u (x, y) = u (x, y) in R n R L a u = 0 in (R n R) \Λ (9) L a u 0 in the sense of distributions in R n R u ττ 0 in R n R, for every tangential unit vector τ where Λ = Λ (u) = {(x, 0) : u (x, 0) = 0}. The following proposition gives a lower bound for the degree k, which implies the optimal regularity of the solution. Lemma 11 If there exists a solution u of problem (9), then k 1 + s = (3 a) /2. Proof. Apply the monotonicity formula in Lemma 10 to w = u τ. Then, L a w = 0 in (R n R) \Λ and, by symmetry, w (x, 0) w y (x, 0) = 0. Moreover, the contact set where w = 0 is convex, since u ττ 0.Therefore w satisfies all the hypotheses of that lemma. Recall that we always assume that (0, 0) F (u) so that w (0, 0) = 0. Thus β (r; w) = 1 y a w (X) 2 r 1 a B r + X n+a 1 dx β (1, w). On the other hand, since w is homogeneous of degree k 2, we have β (r; w) = r2k 2 y a w (X) 2 r 1 a dx = r2k 2 β (1, w). r1 a B + 1 X n+a 1 This implies r 2k 2 r 1 a = r 2s or k 1 + s. From Lemma 11 it would be possible to deduce the optimal regularity of the solution u to (7). However, to study the free boundary regularity we need to classify precisely the solutions to problem (9). For the operator L a, we need to introduce the following subset of the coincidence set. Let { } Λ = (x, 0) R n : lim y 0+ ya u y (x, y) < 0. Notice that Λ is the support of L a u and since L a u = 0 in (R n R) \Λ, we have Λ Λ (Λ is closed). The analysis depend on whether Λ has positive H n measure or not. First examine the case H n (Λ ) = 0. 15

16 Lemma 12 Let u be a solution of problem (9). If H n (Λ ) = 0, then u is a polynomial of degree k. Proof. We know from Lemma 7 that y a u y (x, y) is locally bounded. If H n (Λ ) = 0, then lim y 0 y a u y (x, y) = 0 a.e. x R n. Thus lim y 0 y a u y (x, y) = 0 weak in L and from [23] we infer that u is a global solution of L a u = 0 in R n R. Using Lemma A3 below we conclude the proof. Lemma 13 Let u be a solution of problem (9). If H n (Λ) 0 then, either u 0 or k = 1 + s and Λ is a half n dimensional space. Proof (sketch). First observe that if H n (Λ ) = 0, then u 0, otherwise, from Lemma 12, u (x, 0) would be a polynomial vanishing on a set of positive measure and therefore identically zero. Thus, the polynomial u must have the form u (x, y) = p 1 (x) y p j (x) y 2j and iterating the computation of L a u one deduce p 1 = p 2 =... = p j = 0. Consider now the case H n (Λ ) 0. Then Λ is a thick convex cone. Assume that e n is a direction inside Λ such that a neighborhood of e n is contained in Λ. Using the convexity in the e n direction, we infer that w = u xn cannot be positive at any point X. Moreover, w = 0 on Λ and L a w (X) = 0 in (R n R) \Λ (R n R) \Λ. Thus w must coincide with the first eigenfunction of the weighted spherical Laplacian, minimizer of S 1 θ v 2 y a ds over all v vanishing on Λ and such that S 1 v 2 y a ds = 1. Since Λ is convex, Λ B 1 is contained in half of the sphere B 1 {y = 0}. If it were exactly half of the sphere then it would be given by the first eigenfunction defined in Lemma 9, up to a multiplicative constant, by the explicit expression w (x, y) = ( x 2 n + y 2 x n ) s s = (1 a) /2. On the other hand, the above function is not a solution across {y = 0, x n 0}. Therefore, if Λ B 1 is strictly contained in half of the sphere B 1 {y = 0}, there must be another eigenfunction corresponding to a smaller eigenvalue and consequently to a degree of homogeneity smaller than s. This would imply k < 1 + s, contradicting Lemma 11. The only possibility is therefore k = 1 + s, with Λ = {y = 0, x n 0}. The next theorem gives the classification of asymptotic profiles. Theorem 14 Let u be a non trivial solution of problem (9). There are only two possibilities: 16

17 (1) k = 1 + s, Λ is a half n dimensional space and u depends only on two variables. Up to rotations and multiplicative constants u is unique and there is a unit vector τ such that Λ = {(x, 0) : x τ 0} and ( ) s u τ (x, y) = (x τ) 2 + y 2 (x τ) (2) k is an integer greater than equal to 2, u is a polynomial and H n (Λ) = 0. Proof. If H n (Λ) 0, from Lemma 13 we deduce that, up to rotations and multiplicative constants, there is a unique solution of problem (9), homogeneous of degree k = 1 + s. Moreover, for this solution the free boundary F (u) is flat, that is there is a unit vector (say) e n such that Λ = {(x, 0) : x n 0} and u xn (x, y) = ( x 2 n + y 2 x n ) s. Integrating u xn from F (u) along segments parallel to e n we uniquely determine u (x, 0) = u (x n, 0). If we had another solution v, homogeneous of degree 1 + s, with v (x, 0) = u (x, 0), then necessarily (see the proof of the Liouville-type Lemma A3), for some constant c and y 0, we have v (x, y) u (x, y) = c y s y. But the constant c must be zero, otherwise v u cannot be solution across {y = 0} \Λ. As a consequence, if u is a solution homogeneous of degree 1 + s, with e n normal to F (u), then u = u (x n, y). Indeed, translating in any direction orthogonal to x n and y we get another global solution with the same free boundary. By uniqueness, u must be invariant in those directions. If H n (Λ) = 0, then H n (Λ ) = 0 and from Lemma 12 we conclude that u is a polynomial and k Frequency formula As we have already stated, a crucial tool in order to achieve optimal regularity is given by a frequency formula of Almgren type. If the obstacle were zero, then the frequency formula states that the quantity D a (r; u) = r B r y a u 2 dx B r y a u 2 ds is bounded and monotonically increasing. The conclusion is the following. Theorem 15 Let u be a solution of the zero thin obstacle for the operator L a in B 1. Then D a (r; u) is monotone nondecreasing in r for r < 1. Moreover, D a (r; u) is constant if and only if u is homogeneous. 17

18 When the obstacle ϕ is non zero we cannot reduce to that case. Instead, assuming that ϕ C 2,1, we let ũ (x, y) = u (x, y) ϕ (x) + ϕ (0) 2 (1 + a) y2 so that L a ũ (0) = 0. Moreover Λ = Λ (u) = {ũ = 0}. The function ũ is a solution of the following system: ũ (x, 0) 0 in B 1 where ũ (x, y) = ũ (x, y) in B 1 L a ũ (x, y) = y a g (x) in B 1 \Λ L a ũ (x, y) y a g (x) in B 1, in the sense of distributions g (x) = ϕ (x) ϕ (0) is Lipschitz. Notice that y a g (x) 0 as x 0 and in B 1 \Λ L a ũ (x, y) C y a x. (10) What we expect is a small variation of Almgren s formula. Since u ũ is a C 2,1 function, it is enough to prove any regularity result for ũ instead of u. In order to simplify the notation we will still write u for ũ. Define F (r) = F (r; u) = u 2 y a dσ = r n+a B r (u (rx)) 2 y a ds. B 1 We have: F (r) = = (n + a) r n+a 1 (u (rx)) 2 y a ds + r n+a 2u (rx) u (rx) X y a ds B 1 B 1 = (n + a) r B 1 (u (X)) 2 y a ds + 2u (X) u ν (X) y a ds r B r Thus log F (r) is differentiable for r > 0 and: d dr log F (r) = F (r) F (r) = n + a + r B r 2uu ν y a ds B r u 2 y a ds. Note that the monotonicity of D a (r; u) when ϕ = 0 amounts to say that the function r r d log F (r) = 2D (r; u) + n + a dr is increasing, since in this case 2uu ν y a ds = L a (u 2 ) = ( y a u 2 + 2uL a u)dx B r B r B r = 2 y a u 2 dx. B r 18

19 Due to presence of a nonzero right hand side, we need to prevent the possibility that F (r) become too small under rescaling when compared to the terms involving L a u. It turns out that this can be realized by introducing the following modified formula: Then: Φ (r) = Φ (r; u) = ( r + c 0 r 2) d dr log max [ F (r), r n+a+4]. (11) Theorem 16 (Monotonicity formula). Let u be a solution of problem (10). Then, there exists a small r 0 and a large c 0, both depending only on a, n, ϕ C 2,1, such that Φ (r; u) is monotone nondecreasing for r < r 0. For the proof, we need a Poincaré type estimate and a Rellich type identity. Recall that u (0, 0) = 0 since the origin belongs to the free boundary. Lemma 17 Let u be a solution of problem (10), u (0, 0) = 0. Then (u (X)) 2 y a ds Cr B r u (X) 2 y a dx + c (a, n) r 6+a+n B r and by integrating in r, (u (X)) 2 y a dx Cr 2 B r where c, C depend only on a, n and ϕ C 2,1. Proof. See [22]. B r u (X) 2 y a dx + c (a, n) r 7+a+n Lemma 18 The following identity holds for any r 1. ( ) r θ u 2 u 2 ν y a ds = [(n+a 1) u (X) 2 2 X, u g (X)] y a dx, B r B r where θ u denotes the tangential gradient. Proof. Consider the vector field We have: F = 1 2 ya u 2 X y a X, u u (y > 0). divf = 1 2 (n + a 1)ya u 2 X, u L a u. (12) Since X, u is a continuous function on B r that vanishes on Λ = {u = 0}, we have that X, u L a u has no singular part and coincides with X, u y a g (x). An application of the divergence theorem gives (12). 19

20 Proof of Theorem 16. First we observe that by taking the maximum in (11) it may happen that we get a non differentiable functions. However, max [ F (r), r n+a+4] is absolutely continuous (it belongs to W 1,1 loc (0, 1)) and in any case, the jump in the derivative will be in the positive direction. When F (r) r n+a+4 we have Φ (r) = ( r + c 0 r 2) d log rn+a+4 dr and Φ (r) = (n + a + 4)c 0 > 0. Thus we can concentrate on the case F (r) > r n+a+4 where We have: Φ (r) = ( r + c 0 r 2) d log F (r). dr Φ (r) = ( r + c 0 r 2) B r 2uu ν y a ds B r u 2 y a ds + (1 + c 0r) (n + a) 2Ψ (r) + (1 + c 0 r) (n + a). We show that the first term is increasing, by computing its logarithmic derivative. We find: d dr log Ψ (r) = 1 r + c d c 0 r + dr B r uu ν y a ds B r uu ν y a ds B r 2uu ν y a ds B r u 2 y a ds n + a. r We estimate d dr B r uu ν y a ds from below. Since uu ν y a ds = ( y a u 2 + ul a u)dx B r B r we can write, recalling that L a u c y a x, d uu ν y a ds y a u 2 ds cr (n+a+2)/2 [F (r)] 1/2. dr B r B r We now use Lemma 18 to estimate B r y a u 2 ds from below. We find ( ) y a u 2 ds = u θ 2 + u 2 ν y a ds = B r B r = 2 u 2 ν y a ds + 1 [(n + a 1) u (X) 2 2 X, u g (X)] y a dx = B r r B r = 2 u 2 ν y a ds + n + a 1 uu ν y a ds B r r B r 1 [(n + a 1)u 2 X, u ]g (X) y a dx. r B r 20

21 Therefore d uu ν y a ds 2 u 2 ν y a ds + n + a 1 uu ν y a ds dr B r B r [ r B r ] cr n+a+1 G (r) + r H (r) + rf (r) where G (r) = u 2 y a dx and H (r) = u 2 y a dx. B r B r Collecting all the above estimates, we can write: with and d log Ψ (r) = P (r) + Q (r) dr P (r) = 2 B r u 2 ν y a ds B r uu ν y a ds B r 2uu ν y a ds B r u 2 y a ds 0 Q (r) = c 0 G (r) + rf (r) + r H (r) 1 + c 0 r cr(n+a+1)/2 B r uu ν y a dσ c 0 G (r) + rf (r) + r H (r) 1 + c 0 r cr(n+a+1)/2 H (r) r (n+a+2)/2. G (r) First we estimate F, G, H. Since F (r) > r n+4+a, from the Poincaré Lemma 17 we have: r n+4+a < F (r) CrH (r) + c (a, n) r 6+a+n. Integrating the above inequalities in r, we get: G (r) = r 0 F (s) ds Cr 2 H (r) + c (a, n) r 7+a+n. This means that, for small enough r 0 and r < r 0 : F (r) crh (r) and G (r) Cr 2 H (r) so that: Q (r) c cr r H (r) cr(n+a+1)/2 H (r) r (n+a+4)/2. H r Since r n+4+a < F (r) CrH (r), we also have H (r) cr n+a+3 and for r 0 small: Q (r) c c 0 r r cr(n+a+1)/2 H (r) r (n+a+4)/2 c 0 n+a+1 cr 2 +1 n+a+3 2 = c c 0 r 1 + cr c which is positive if c 0 is large and r 0 small. 21

22 2.7 Blow-up sequences and optimal regularity The optimal regularity of the solution can be obtained by a careful analysis of the possible values of Φ (0+). When Φ is constant and the obstacle is zero, [Φ (0+) n a] /2 represents the degree of homogeneity at the origin. Thus, by a suitable blow-up of the solution, we will be able to classify the possible asymptotic behaviors at the origin, using the results of Section 2.5. The first result is the following. Theorem 19 Let u be a solution of problem (10), u (0, 0) = 0. Then either Φ (0+; u) = n + a + 2 (1 + s) or Φ (0+; u) n + a + 4. To prove the theorem, guided by the zero obstacle case in [10], a key point is to introduce the following rescaling: where u r (X) = u (rx) d r (13) ) 1/2 ( 1/2 d r = (r (n+a) u 2 y a dσ = r (n+a) F (r)). B r Notice that the natural rescaling u (rx) /r µ, where µ = Φ (0+; u) n a, would not be appropriate, because on this kind of rescaling we have precise control of its behavior as r 0 merely from one-side. Rescaling by an average over smaller and smaller balls provides the necessary adjustments for controlling the oscillations of u around the origin.two things can occur: d r lim inf r 0 r 2 { = + first case < + second case. (14) The next lemma takes care of the first case. Lemma 20 Let u be as in Theorem 19. Assume that d r lim inf r 0 r 2 = +. Then, there is a sequence r k 0 and a function U 0 : R n+1 R, non identically zero, such that: 1. u rk U 0 in W 1,2 ( B 1/2, y a ) 2. u rk U 0 uniformly in B 1/2. 3. x u rk x U 0 uniformly in B 1/2. 4. y a y u rk y a y U 0 uniformly in B 1/2. 22

23 Moreover, U 0 is a solution of system (9) and its degree of homogeneity is [Φ (0+; u) n a]/2. Proof. First of all, observe that u r L 2 ( B 1, y a ) = 1 for every r. Using the frequency formula, the Poincaré type Lemma A5 below and the semiconvexity of u r, one can show that u r is bounded in W ( 1,2 B 1/2, y a ). Now, u + r and u r are subsolutions of the equation L a w cr y a x and therefore (see [29]) u r is bounded in L ( B 3/4 ). The semiconvexity of u in the x variable gives, for every tangential direction τ, ττ u r = r2 d r u ττ (rx) C r2 d r. (15) From Lemma 8 we obtain a subsequence u rk strongly convergent in W 1,2 ( B 1/2, y a ) to some function U 0 as r k 0. On the other hand (Theorem 16) we know that Φ (r; u) is monotone and converges to Φ (0+; u) as r 0. We have: Φ (rs; u) rs B rs u (X) 2 y a dx B rs u 2 y a ds + (n + a) = r B r u s (X) 2 y a dx B r u 2 s y a ds We want to set s = r k 0 and pass to the limit in the above expression to obtain: Φ (0+; u) (n + a) = r B r U 0 (X) 2 y a dx B r U0 2. (16) y a ds This is possible since one can show that B r u 2 s y a ds c > 0. From (15), we have ττ u rk C r2 k d rk 0 and therefore U 0 is tangentially convex. On the other hand, each u r satisfies the following conditions: (a) u r (x, 0) 0 in B 1. (b) (c) Since L a u r = r2 a L a u (rx) = r2 y a g (rx) in B 1 \Λ (u r ) d r d r L a u r r2 d r y a g (rx) in B 1. r 2 d r y a g (rx) c r2 d r y a r x 0 as r (n + a).

24 it follows that U 0 is a solution of the homogeneous problem U 0 (x, 0) 0 in B 1 U 0 (x, y) = U 0 (x, y) in B 1 L a U 0 = 0 in B 1 \Λ L a U 0 0 in B 1, in the sense of distributions For this problem, the frequency formula holds without any error correction. Thus we conclude that U 0 is homogeneous in B 1/2 and its degree of homogeneity is exactly (Φ (0+; u) (n + a)) /2. Since it is homogeneous, then it can be extended to R n+1 as a global solution of the homogeneous problem. Finally, from the a priori estimates in Lemma 7, it follows that we can choose r k so that the sequences u rk, u rk and y a u rk converge uniformly in B 1/2. Proof of Theorem 19. In the first case of (14), we use Lemma 20 and Theorem 14 to find the blow-up profile U 0 and to obtain that the degree of homogeneity of U 0 is 1 + s or at least 2. Therefore Φ (0+; u) = Φ (0+; U 0 ) = n+a+2 (1 + s) or Φ (0+; u) = Φ (0+; U 0 ) n+a+4. Consider now the second case of (14). If F (r k ; u) < r n+a+4 k for some sequence r k 0, then, for these values of r k, Φ (r k ; u) = (n + a + 4) (1 + c 0 r k ) so that Φ (0+; U 0 ) = n + a + 4. On the other hand, assume that F (r; u) r n+a+4 for r small. Since we are in the second case, for some sequence r j 0 we have d rj /rj 2 C so that r n+a+4 j Taking logs in the last inequality, we get F (r j ; u) Cr n+a+4 j. (n + a + 4) log r j log F (r j ; u) C + (n + a + 4) log r j. We want to show that Φ (0+; u) n + a + 4. By contradiction, assume that for small r j we have Φ (r j ; u) n + a + 4 ε 0. Take r m < r j 1 and write: (n + a + 4) (log r j log r m ) C log F (r j ; u) log F (r m ; u) rj d rj ( = log F (s; u) ds r + c0 r 2) rj 1 Φ (s; u) ds r 1 Φ (r j ; u) ds r m dr r m r m (n + a + 4 ε 0 ) (log r j log r m ) which gives a contradiction if we make (log r j log r m ) +. From the classification of the homogeneity of a global profile, we may proceed to identify the unique limiting profile U 0, along the whole sequence u r, modulus rotations. precisely, we have: 24

25 Proposition 21 Let u be as in Theorem 19. Assume that Φ (0+; u) = n + a + 2 (1 + s). There is a family of rotations A r, with respect to x, such that u r A r converges to the unique profile U 0 of homogeneity degree 1 + s. More precisely: 1. u r A r U 0 in W 1,2 ( B 1/2, y a ). 2. u r A r U 0 uniformly in B 1/2. 3. x (u r A r ) x U 0 uniformly in B 1/2. 4. y a y (u r A r ) y a y U 0 uniformly in B 1/2. We are now ready to prove the optimal 1 + s decay of u at (0, 0). This is done in two steps. First, we control the decay of u in terms of the decay of F (r; u). In turn, the frequency formula provides the precise control of F (r) from above. This is the content of the next two lemmas. Lemma 22 If F (r; u) cr n+a+2(1+α) (17) for every r < 1, then u (0, 0) = 0, u (0, 0) = 0 and u is C 1,α at the origin in the sense that u (X) C 1 X 1+α for X 1/2 and C 1 = C 1 (C, n, a). Proof. Consider u + and u, the positive and negative parts of u. We have already noticed that L a u +, L a u C y a x. For some r > 0, let h be the L a harmonic replacement of u + in B r. Note that ( ) 0 = L a h L a u + + C X 2 r 2. 2 (n + a + 1) Hence, by comparison principle h u + C r 2. From (17) we have: h 2 y a dσ = (u + ) 2 y a dσ cr n+a+2(1+α). B r B r Since w (X) = y a is a A 2 weight, from the local L estimates (see [29]) we conclude sup B r/2 h (X) C 1 r 1+α. Then u + h + r 2 Cr 1+α in B r/2. A similar estimate holds for u and the proof is complete. 25

26 Lemma 23 If Φ (0+; u) = µ then for any r < 1 and c = c (F (1; u), c 0 ). F (r; u) cr µ Proof. Let f (r) = max { F (r), r n+a+4} F (r). Since Φ is nondecreasing: and then An integration gives: µ = Φ (0+) Φ (r) = ( r + c 0 r 2) d log f (r) dr d dr log f (r) µ (r + c 0 r 2 ). log f (1) log f (r) 1 µ r (s + c 0 s 2 ) ds 1 [ ] 1 µ log r µ r s µ (s + c 0 s 2 ds ) µ log r C 1 µ so that log f (r) log f (1) + µ log r + C 1 µ. Taking the exponential of the two sides we infer with C = f (1) e C1µ. F (r) f (r) Cr µ The optimal decay of a solution u of (10) follows easily. Precisely: Theorem 24 Let u be a solution of (10), with u (0, 0) = 0. Then u (X) C X 1+s sup u, B 1 ) where C = C (n, a, g Lip. Proof. From Theorem 19, µ = Φ (0+) n + a + 2 (1 + s) and the conclusion follows from Lemmas 22 and 23. The optimal regularity of the solution of the obstacle problem for the fractional laplacian is now a simple corollary. Corollary 25 (Optimal regularity of solutions) Let ϕ C 2,1. Then the solution u of the obstacle problem for the operator ( ) s belongs to C 1,s (R n ). 26

27 Proof. Using the equivalence between the obstacle problem for ( ) s and the thin obstacle for L a, Theorem 24 shows that u ϕ has the right decay at free boundary points. This is enough to prove that u C 1,s (R n ). Remark 2 Observe that it is not true that the solution of the thin obstacle problem for L a is C 1,s in both variables x and y. It is quite interesting however, that the optimal decay takes place in both variables at a free boundary point. In any case, we have that a solution u of one of the systems (7) or (10) belongs to C 1,s (B 1/2 ), for every y 0 (0, 1/2). 2.8 Nondegenerate case. Lipschitz continuity of the free boundary In analogy with what happens in the zero obstacle problem, the regularity of the free boundary can be inferred for points around which u has an asymptotic profile corresponding to the optimal homogeneity degree Φ (0+; u) = n + a + 2 (1 + s). Accordingly, we say that X 0 F (u) is regular or stable if µ (X 0 ) = Φ (0+; u) = n + a + 2 (1 + s). As always, we refer to the origin (X 0 = (0, 0)). The strategy to prove regularity of F (u) follows the well known pattern first introduced by Athanasopoulos and Caffarelli in [6] and further developed by Caffarelli in [18]. The first step is to prove that in a neighborhood of (0, 0) there is a cone of tangential directions (cone of monotonicity) along which the derivatives of u are nonnegative and have nontrivial growth. In particular, the free boundary is the graph x n = f (x 1,..., x n 1 ) of a Lipschitz function f.the second step is to prove a boundary Harnack principle assuring the Hölder continuity of the quotient of two nonnegative tangential derivatives. This implies that F (u) is locally a (n 2) dimensional manifold of class C 1,α. The following theorem establishes the existence of a cone of monotonicity. Theorem 26 Assume µ < n + a + 4. Then, there exists a neighborhood B ρ of the origin and a tangential cone Γ (θ, e n ) R n {0} such that, for every τ Γ (θ, e n ), we have τ u 0 in B ρ. In particular, the free boundary is the graph x n = f (x 1,..., x n 1 ) of a Lipschitz function f. The theorem follows by applying the following lemma to a tangential derivative h = τ u r, where u r is the blow-up family (13) that defines the limiting profile U 0, for r small. Lemma 27 Let Λ be a subset of R n {0}. Assume h is a continuous function with the following properties: 1. L a h γ y a in B 1 \Λ. 2. h 0 for y σ > 0, h = 0 on Λ. 27

28 3. h c 0 for y 1 + a/8n. 4. h ω (σ) for y < σ, where ω is the modulus of continuity of h. There exist σ 0 = σ 0 (n, a, c 0, ω) and γ 0 = γ 0 (n, a, c 0, ω) such that, if σ < σ 0 and γ < γ 0, then h 0 in B 1/2. Proof. Suppose by contradiction X 0 = (x 0, y 0 ) B 1/2 and h (X 0 ) < 0. Let Q = and P (x, y) = x x 0 2 Then: { (x, y) : x x 0 < 1 } 1 + a 3, y 4n n a+1 y2. Observe that L a P = 0. Define v (X) = h (X) + δp (X) v (X 0 ) = h (X 0 ) + δp (X 0 ) v (X) 0 on Λ γ 2(a+1) y2 0 < 0 L a v = L a h + δl a P γ y a 0 outside Λ. γ 2(a + 1) y2. Thus, v must have a negative minimum on Q. On the other hand, if δ, γ are small enough, then v 0 on Q and we have a contradiction. Therefore h 0 in B 1/2. Proof of Theorem 26. Since µ = Φ (0; u) < n + a + 4, Theorem 19 gives µ = n + a + 2 (1 + s). Moreover, the blow-up sequence u r converges (modulus subsequences) to the global profile U 0, whose homogeneity degree is 1 + s and whose free boundary is flat. Let us assume that e n is the normal to the free boundary of U 0. Then n U 0 (x, y) = c ( x 2 n + y 2 x n ) s. For some θ 0 > 0, let σ any vector orthogonal to y and x n such that σ < θ 0. From Theorem 15 we know that U 0 is constant in the direction of σ and therefore, if τ = e n + σ, τ U 0 = n U 0. On the other hand, x u r x U 0 uniformly in every compact subset of R n+1. Thus, for every δ 0, there is an r for which τ U 0 τ u r δ 0 where τ = e n + σ. If we differentiate the equation L a u r (X) = r2 d r y a g (rx) 28

29 we get L a [ τ u r (X)] = r2 d r y a r τ g (rx) Cr y a in B 1 \Λ (u r ) (18) and the right hand side tends to zero as r 0. Thus, for r small enough, τ u r satisfies all the hypotheses of Lemma 27 and therefore is nonnegative in B 1/2. This implies that near the origin, the free boundary is a Lipschitz graph. 2.9 Boundary Harnack principles and C 1,α regularity of the free boundary Growth control for tangential derivatives We continue to examine the regularity of F (u) at stable points. As we have seen, at these point we have an exact asymptotic picture and this fact allows us to get a minimal growth for any tangential derivative when u r is close to the blow-up limit u 0. This is needed in extending the Carleson estimate and the Boundary Harnack principle in our non-homogeneous setting. First, we need to refine Lemma 27. Lemma 28 Let δ 0 = (12n) 1/2s. There exists ε 0 = ε 0 (n, a) > 0 such that if v is a function satisfying the following properties: 1. L a v (X) ε 0 for X B 1 (0, δ 0 ), 2. v (X) 0 for X B 1 (0, δ 0 ), 3. v (x, δ 0 ) 1 4n for x B 1, then Proof. Compare v with v (x, y) C y 2s in B 1/2 [0, δ 0]. ( w (x, y) = 1 + ε ) 0 y 2 x x y 2s. 2 n Inside B 1 (0, δ 0 ), to show that w v in B 1 (0, δ 0 ). Corollary 29 Let u be a solution of (10) with g, g ε 0. Let u 0 the usual asymptotic nondegenerate profile and assume that x u u 0 ε 0. Then, if ε 0 is small enough, there exists c = c (n, a) such that u τ (X) c dist (X, Λ) 2s for every X B 1/2 and every tangential direction τ such that τ e n < 1/2. 29

30 Proof. From (18) and Theorem 26, we know that u τ is positive in B 1/2. Applying Lemma 28 we get u τ c y 2s in B 1/4. Let now X = (x, y) B 1/8 and d =dist(x, Λ). Consider the ball B d/2 (X). At the top point of this ball, say (x T, y T ) we have y T d/2. Therefore By Harnack s inequality, Boundary Harnack u τ (x T, y T ) cd 2s. u τ (x, y) cu τ (x T, y T ) cd 2s. Using the growth control from below provided by Lemma 28 it is easy to extend the Carleson estimate to our nonhomogeneous setting. Lemma 30 (Carleson estimate). Let D = B 1 \Λ where Λ {y = 0}. Assume that Λ B 1 is given by a Lipschitz (n 2) manifold with Lipschitz constant L. Let w 0 in D, vanishing on Λ. Assume in addition that: 1. L a w c y a in D; 2. nondegeneracy: w (X) Cd β X for some β (0, 2), where d X =dist(x, Λ). Then, for every Q Λ B 1/2 and r small: sup w C (n, a, L) w (A r (Q)), B r(q) D where A r (Q) is a point such that B ηr (A r (Q)) B r (Q) D for some η depending only on n and L. Proof. Let w be the harmonic replacement of w in B 2r (Q) D, r small. Standard arguments (see e.g. [21]) give w (X) Cw (A r (Q)) in B r (Q) D. ) On the other hand, comparing w with the function w (X)+C ( X Q 2 r 2 we get w w cr 2 in B 2r (Q) D. Thus w (X) C [ w (A r (Q)) + cr 2] in B r (Q) D. 30

31 From the nondegeneracy condition we infer w (A r (Q)) cr β and since β < 2, the theorem follows. The following theorem expresses a boundary Harnack principle valid in our nonhomogeneous setting. Theorem 31 (Boundary Harnack principle). Let D = B 1 \Λ where Λ {y = 0}. Let v, w positive functions in D satisfying the hypotheses 1 and 2 of Lemma 30 and symmetric in y. Then there is a constant c = c (n, a, L) such that v (X) w (X) c v ( ) 0, 1 2 w ( ) 0, 1 in B 1/2. 2 Moreover, the ratio v/w is Hölder continuous in B 1/2, uniformly up to Λ. Proof. Let us normalize v, w setting v ( 0, 1 2) = w ( 0, 1 2) = 1. From the Carleson estimate and Harnack inequality, for any δ > 0 we get: and v (X) C in B 3/4 w (X) c in B 3/4 { y > δ}. This implies that, for a constant s small enough, v sw fulfills the conditions of Lemma 26. Therefore v sw 0 in B 1/2 or, in other words: v (X) w (X) s in B 1/2. At this point, the rest of proof follows by standard iteration C 1,α regularity of the free boundary As in the case of the thin-obstacle for the Laplace operator, the C 1,α regularity of the free boundary follows by applying Theorem 31 to the quotient of two positive tangential derivatives. Precisely we have: Theorem 32 Let u be a solution of (10). Assume ϕ C 2,1 and Φ (0) < n+a+4. Then the free boundary is a C 1,α (n 1) dimensional surface around the origin. Remark 3 As we have already noted, the boundary Harnack principle in Theorem 31 is somewhat weaker than the usual one. Notice the less than quadratic decay to zero of the solution at the boundary, necessary to control the effect of the right hand side. 31

32 2.10 Non regular points on the free boundary As we have seen, the regularity of the free boundary can be achieved around regular or stable points, corresponding to the optimal homogeneity [Φ (0+; u) n (a + 4)] /2 = 1 + s. On the other hand, there are solutions of the zero-thin obstacle problem like ρ k+1/2 cos 2k + 1 θ or ρ 2k cos 2kθ, k 2, 2 vanishing of higher order at the origin. In these cases we cannot expect any regularity of the free boundary. The non regular points of the free boundary can be divided in two classes: the set Σ (u) at which Λ (u) has a vanishing density (singular points), that is { H n (Λ (u) B } r (x 0 )) Σ (u) = (x 0, 0) F (u) : lim r 0 + r n = 0, and the set of non regular, non singular points. [36]), given by the harmonic polynomial The following example (see p (x 1, x 2, y) = x 2 1x 2 2 (x x 2 2)y y4 shows that the entire free boundary of the zero-thin obstacle problem (2) could be composed by singular points. In fact F (p) = Λ (p) is given by the union of the lines x 1 = y = 0 and x 2 = y = 0. We shall see that, as in this example, the singular set is contained in the union of C 1 manifold of suitable dimension ([36]) Structure of the singular set (zero thin obstacle) In this section we describe the main results and ideas from ([36]). Their techniques works also for the fractional Laplacian but, for the sake of simplicity, we present them in the (important case) of the thin obstacle problem. Consider the following problem: u ϕ 0 in B 1 u = 0 in B 1 \ {(x, 0) : u (x, 0) = ϕ (x)} (u ϕ)u xn = 0, u xn 0 in B 1 u (x, y) = u (x, y) in B 1. where ϕ : B R. For better clarity we outline the proofs in the special case ϕ = 0. As we shall see, around the singular points a precise analysis of the behavior of u and the structure of the free boundary can be carried out. The analysis of the free boundary around the other kind of points is still an open question in general. However, in some important special cases, complete information can be given, as we shall see in the sequel. 32

33 It is convenient to classify a point on F (u) according to the degree of homogeneity of u, given by the frequency formula centered at that point. In other words, set and define Φ X0 (r; u) = r B r(x 0) u 2 B r(x 0) u2 ds F κ (u) = { X 0 F (u) : Φ X0 (0+; u) = κ }, Σ κ (u) = Σ (u) F κ (u). According to these notations, X 0 is a regular point if it belongs to F 3/2 (u). Since r Φ X0 (r; u) is nondecreasing, it follows that the mapping X 0 Φ X0 (0+; u) is upper-semicontinuous. Moreover, since Φ X0 (0+; u) misses all the values in the interval (3/2, 2), it follows that F 3/2 (u) is a relatively open subset of F (u). Before stating the structure theorems of Σ (u), it is necessary to examine the asymptotic profiles obtained at a singular point from the rescalings v r (X) = v (rx) /(r n B r(x 0) u2 ) 1/2 ; indeed, saying that X 0 = (x 0, y) Σ (u) is equivalent to state that lim H n (Λ (v r ) B r (x 0 )) = 0. (19) As we see immediately, this implies that any blow-up v at a singular point is harmonic in B 1 (X 0 ). Moreover, it is possible to give a complete characterization of these blow-ups in terms of the value κ = Φ X0 (r; u). In particular Σ κ (u) = F κ (u) for κ = 2m, m N. Theorem 33 (Blow-ups at singular points). Let (0, 0) F κ (u). The following statements are equivalent: (i) (0, 0) Σ κ (u). (ii) Any blow-up of u at the origin is a non zero homogeneous polynomial p κ of degree κ satisfying p κ = 0, p κ (x, 0) 0, p κ (x, y) = p κ (x, y). (iii) κ = 2m for some m N. Proof. (i) = (ii). Since u is harmonic in B 1 ±, we have: v r = 2( y v r )H n Λ(v r) in D (B 1 ). (20) Then v r is equibounded in H 1 loc (B 1), and (19) says that H n (Λ (v r ) B 1) 0 33

34 as r 0. Thus (20) implies that v r 0 in D (B 1 ), and therefore any blow-up v must be harmonic in B 1. From Section 2.7 we know that v is homogeneous and non trivial, and thus it can be extended to a harmonic function in all of R n+1. Being homogeneous, v has at most a polynomial growth at infinity, hence Liouville Theorem implies that v is a non trivial homogeneous harmonic polynomial p κ of integer degree κ. The properties of u imply that p κ (x, 0) 0, and p κ (x, y) = p κ (x, y) in R n+1. (ii) = (iii). We must show that κ is an even integer. If κ is odd, the nonnegativity of p κ on y = 0 implies that p κ vanishes on the hyperplane y = 0. On the other hand, from the even symmetry in y we infer that y p κ (x, 0) 0 in R n. Since p κ is harmonic, the Cauchy-Kowaleskaya Theorem implies that p κ 0 in R n+1. Thus κ = 2m, for some m N. (ii) = (i). Suppose (0, 0) is not a singular point. sequence r j 0 such that H n (Λ (v r ) B 1) δ > 0. Then, there exists a We may assume that v rj converges to a blow-up p. We claim that H n (Λ (p ) B 1) δ > 0. Indeed, otherwise, there exists an open set U R n with H n (U) < δ such that Λ (p ) B 1 U. Then, for j large, we must have Λ (v r ) B 1 U which is a contradiction, since H n (Λ ( ) v rj B 1) δ > H n (U). This implies that p (x, 0) 0 in R n and consequently in R n+1, by the Cauchy-Kowaleskaya theorem. Contradiction to (ii). (iii) = (ii). From Almgren s formula, any blow-up is a κ homogeneous solution of the zero thin obstacle problem in R n+1. Then v = 2v y H Λ(v) n in R n+1, with v y 0 on y = 0. Since κ = 2m, the following auxiliary lemma implies that v = 0 in R n+1 and therefore v is a polynomial. Lemma 34 Let v H 1 loc ( R n+1 ) satisfy v 0 in R n+1 and v = 0 in R n+1 \ {y = 0}. If v is homogeneous of degree κ = 2m then v = 0 in R n+1. Proof. By assumption, µ = v is a nonpositive measure, supported on {y = 0}. We have to show that µ = 0. Let q be a 2m homogeneous harmonic polynomial, which is positive on {y = 0} \ (0, 0). For instance: q (X) = n Re(x j + iy) 2m. j=1 Take ψ C 0 (0, + ) such that ψ 0 and let Ψ (X) = ψ ( X ). Then, we have: 34

35 µ, Ψq = v, Ψq = (Ψ v q + q v Ψ) dx R n+1 = ( Ψv q v q Ψ + q v Ψ) dx R n+1 = [ Ψv q v ψ ( X ) R X n+1 = 0 (X q) + q ψ ( X ) (X v)]dx X since q = 0, X q = 2mq, X v = 2mv. This implies that µ is supported at X = 0, that is µ = cδ (0,0). On the other hand, δ (0,0) is homogeneous of degree (n + 1) while µ is homogeneous of degree 2m 2 and therefore µ = 0. Definition 1 We denote by P κ the class of homogeneous harmonic polynomials of degree κ = 2m defined in Theorem 33, that is: P κ = {p κ : p κ = 0, p κ X = κp κ, p κ (x, 0) 0, p κ (x, y) = p κ (x, y)}. (21) Via the Cauchy-Kovaleskaya Theorem, it is easily shown that the polynomials in P κ can be uniquely determined from their restriction to the hyperplane y = 0. Thus, if p κ P κ is not trivial, then also its restriction to y = 0 must be non trivial. The next theorem gives an exact asymptotic behavior of u near a point X 0 Σ κ (u). Theorem 35 (κ differentiability at singular points) Let X 0 Σ κ (u), with κ = 2m, m N. Then there exists a non trivial p X0 κ P κ such that u (X) = p X0 κ (X X 0 ) + o ( X X 0 κ ). (22) Moreover, the mapping X 0 p X0 κ is continuous on Σ κ (u). The proof is given in Subsection B... Note that, since P κ is a convex subset of the finite dimensional space of the homogeneous polynomial of degree κ, all the norms on P κ are equivalent. Thus, the continuity in Theorem 35 can be understood, for instance, in the L 2 ( B 1 ) norm. The structure of F (u) around a singular point X 0 depends on the dimension of the singular set at that point, as defined below in terms of the polynomial p X0 κ : Definition 2 (Dimension at a singular point) Let X 0 Σ κ (u). The dimension of Σ κ (u) at X 0 is defined as the integer Since p X0 κ d X0 κ = dim { ξ R n : ξ x p X0 κ (x, 0) = 0 for all x R n}. (x, 0) is not identically zero on R n, we have For d = 0, 1,..., n 1 we define 0 d X0 κ n 1. Σ d κ (u) = { X 0 Σ κ (u) : d X0 κ = d }. 35

36 x 2 Σ 1 2 Σ 1 2 (0, 0) x 1 Σ 0 4 Σ 1 2 Σ 1 2 Figure 1: Free boundary for u(x 1, x 2, y) = x 2 1x 2 2 ( x x 2 2) y y4 in R 3 with zero thin obstacle on R 2 {0}. (Credit: Garofalo-Petrosyan, 2009) Here is the structure theorem: Theorem 36 (Structure of the singular set) Every set Σ d κ (u), κ = 2m, m N, d = 0, 1,..., n 1, is contained in a countable union of d dimensional C 1 manifolds. For the harmonic polynomial p (x 1, x 2, y) = x 2 1x 2 2 (x x 2 2)y y4 considered above, it is easy to check that (0, 0) Σ 0 4 (u) and the rest of the points on F (u) belongs to Σ 1 2 (u) (see Figure 1). As in the classical obstacle problem, the main difficulty in the analysis consists in establishing the uniqueness of the Taylor expansion (22), which in turn is equivalent to establish the uniqueness of the limiting profile obtained by the sequence of rescalings u r. A couple of monotonicity formulas, strictly related to Almgren s formula and to formulas in [65] and [53], play a crucial role in circumventing these difficulties Monotonicity formulas We introduce here two main tools. We start with a one-parameter family of monotonicity formulas (see also citew) based on the functional: 1 Wκ X0 (r; u) = r n 1+2κ u 2 dx κ r n+2κ u 2 ds B r(x 0) 1 H (r) κ K (r). rn 1+2κ rn+2κ 36 B r(x 0)

37 where κ 0. If X 0 = (0, 0) we simply write W κ (r; u). The functionals Wκ X0 (r; u) and Φ X0 (r; u) are strictly related. Indeed, taking for brevity X 0 = (0, 0), we have: W κ (r; u) = K (r) [Φ (r; u) κ]. (23) rn+2κ This formula shows that Wκ X0 (r; u) is particularly suited for the analysis of asymptotic profiles at points X 0 F κ (u). Moreover, for these points, since from Almgren s frequency formula we have Φ X0 (r; u) Φ X0 (0+; u) = κ, we deduce that W X0 κ (r; u) 0. (24) The next theorem shows the main properties of W κ (r; u). Theorem 37 (W type monotonicity formula) Let u be a solution of our zero obstacle problem in B 1. Then, for 0 < r < 1, d dr W κ (r; u) = 1 r n+2κ (X u κu) 2 ds. B r As a consequence, the function r W κ (r; u) is nondecreasing in (0, 1). Moreover, W κ ( ; u) is constant if and only if u is homogeneous of degree κ. and Proof. Using the identities H (r) = u 2 ds and K (r) = n B r r K (r) + 2 uu ν ds (25) B r we get: = = = B r u 2 ds = (n 1) r u B r (u ν ) 2 ds. B r (26) d dr W κ (r; u) { 1 r n 1+2κ H (r) n 1 + 2κ V (r) κ } κ (n + 2κ) r r K (r) + r 2 K (r) { 1 r n 1+2κ 2 u 2 νds 4κ } uu ν ds + B r r Br 2κ2 r 2 u 2 ds B r 2 r n+1+2κ (X u κu) 2 ds. B r The next one is a generalization of a formula in [53], based on the functional Mκ X0 (r; u, p κ ) = 1 r n+2κ (u p κ ) 2 ds. 37 B r(x 0)

38 We set M κ (r; u, p κ ) = M κ (0,0) (r; u, p κ ). Here κ = 2m and p κ is a polynomial in the class P κ defined in (21). Since it measures the distance of u from an homogeneous polynomial of even degree, it is apparent that Mκ X0 (r; u, p κ ) is particularly suited for the analysis of blow-up profiles at points X 0 Σ κ (u). We have: Theorem 38 (M type monotonicity formula) Let u be a solution of our zero obstacle problem in B 1. Assume (0, 0) Σ κ (u), κ = 2m, m N. Then, for 0 < r < 1, the function r M κ (r; u, p κ ) is nondecreasing in (0, 1). Proof. We show that Let w = u p κ. We have: d 1 dr r n+2κ = 1 r 2κ+1 d dr M κ (r; u, p κ ) 2 r W κ (r; u) 0. B r w 2 ds = d dr 1 r 2κ w (ry ) 2 ds B 1 [w (ry ) [ w (ry ) ry κw(ry )]ds B 1 2 = r n+1+2κ w(x w κw)ds B r On the other hand, since Φ (r; p κ ) = κ, it follows that W κ (r; p κ ) = 0 and we can write: W κ (r; u) = W κ (r; u) W κ (r; p κ ) 2 = r n 1+2κ ( w w p κ )dx κ B r r n+2κ (w 2 + 2wp κ )ds B r 2 = r n 1+2κ w 2 dx κ B r r n+2κ w 2 ds + w(x p κ κp κ )ds B r B r 2 = r n 1+2κ w 2 dx κ B r r n+2κ w 2 ds B r 2 = r n 1+2κ w wdx + 1 B r r n+2κ w(x w κw)ds B r 1 r n+2κ w(x w κw)ds = r d B r 2 dr M κ (r; u, p κ ) since w w = (u p κ ) ( u p κ ) = p κ u 0. 38

39 Proofs of Theorems 35 and 36 With the above monotonicity formulas at hands we are ready to prove Theorems 35 and 36. Recall that, from the frequency formula, we have the estimate u (X) c X κ in B 2/3 (27) for any solution of our free boundary problem. At a singular point, we have also a control from below. Lemma 39 Let u be a solution of our zero obstacle problem in B 1. Assume (0, 0) Σ κ (u). Then, sup u cr κ (0 < r < 2/3). (28) B r Proof. Suppose that (28) does not hold. Then, for a sequence r j 0 we have ( ) 1/2 h j = u 2 ds = o ( r κ ) j. B rj We may also assume that (see Lemma 33) v j (X) = u (r jx) h j q κ (X) uniformly on B 1, for some q κ P κ. Since B 1 qκds 2 = 1, it follows that q κ is non trivial. Under our hypotheses, we have 1 M κ (0+; u, q κ ) = lim j r n+2κ j (u q κ ) 2 ds = B rj B 1 q 2 κds = 1 r n+2κ j B rj q 2 κds. Hence, or Rescaling, we obtain (u q κ ) 2 ds qκds 2 B rj B rj h j r κ j B rj (u 2 2uq κ )ds 0. B rj ( h j r κ j v 2 j 2v j q κ )ds 0. Dividing by h j r κ j and letting j we get qκds 2 0 B rj 39

40 which gives a contradiction, since q κ is non trivial. Given the estimates (27) and (28) around a point X 0 Σ κ (u), it is natural to introduce the family of homogeneous rescalings given by u (κ) r (X) = u (rx + X 0) r κ. From the estimate (27) we have that, along a sequence r = r j, u κ r u 0 in C 1,α loc (Rn ). We call u 0 homogeneous blow-up. Lemma 39 assures that u 0 is non trivial. The next results establishes the uniqueness of these asymptotic profiles and proves the first part of Theorem 35. Theorem 40 (Uniqueness of homogeneous blow-up at singular points) Assume (0, 0) Σ κ (u). Then there exists a unique non trivial p κ P κ such that As a consequence, (22) holds. u (κ) r (X) = u (rx) r κ p κ (X). Proof. Consider a homogeneous blow-up u 0. For any r > 0 we have: W κ (r; u 0 ) = lim W rj 0 κ(r; u (κ) j ) = lim W rj 0 κ(rr j ; u) = lim W rj 0 κ(0+; u). From Theorem 35 we infer that u 0 is homogeneous of degree κ. The same arguments in the proof of Lemma 34 give that u 0 must be a polynomial p κ P κ. To prove the uniqueness of u 0, apply the M monotonicity formula to u and u 0. We have: M κ (0+; u, u 0 ) = c n lim (u (κ) j u 0 ) 2 ds = 0. j B 1 In particular, by monotonicity, we obtain also that c n (u (κ) r u 0 ) 2 ds = M κ (r; u, u 0 ) 0 B 1 as r 0, and not just over a subsequence r j. Thus, if u 0 is a homogeneous blow-up, obtained over another sequence r j 0, we deduce that (u 0 u 0 ) 2 ds = 0. B 1 Since u 0 and u 0 are both homogeneous of degree κ, they must coincide in R n+1. The next lemma gives the second part of Theorem

41 Lemma 41 (Continuous dependence of the blow-ups) For X 0 Σ κ (u) denote by p X0 κ the blow-up of u obtained in Theorem 40 so that: u (X) = p X0 κ (X X 0 ) + o ( X X 0 κ ). Then, the mapping X 0 p X0 κ from Σ κ (u) to P κ is continuous. Moreover, for any compact K Σ κ (u) B 1, there exists a modulus of continuity σ K, σ K (0+) = 0, such that u (X) p X 0 κ (X X 0 ) σk ( X X 0 ) X X 0 κ for every X 0 K. Proof. As we have already observed, we endow P κ with the L 2 ( B 1 ) norm. The first part of the lemma follows as in Theorem 40. Indeed, fix ε > 0 and r ε such that Mκ X0 (r ε ; u, p X0 κ ) = 1 rε n+2κ B rε (u (X + X 0 ) p X0 κ ) 2 ds < ε. There exists δ ε such that if X 0 Σ κ (u) and X 0 X 0 < δ ε, then M X 0 κ (r ε ; u, p X0 κ ) = 1 rε n+2κ (u (X + X 0) p X0 κ ) 2 ds < 2ε. B rε By monotonicity, we deduce that, for 0 < r < r ε, M X 0 κ (r; u, p X0 κ ) = 1 r n+2κ (u (X + X 0) p X0 κ ) 2 ds < 2ε. B r Letting r 0 we obtain M X 0 κ (0+; u, p X0 κ ) = c n 0 (p X κ B 1 p X0 κ ) 2 ds < 2ε and therefore the first part of the lemma is proved. To show the second part, note that if X 0 X 0 < δ ε and 0 < r < r ε, we have: u ( + X 0) p X 0 κ L 2 ( B r) u ( + X 0 ) p X0 p κ L 2 ( B + X 0 r) κ p X 0 κ 2 (2ε) 1/2 r κ+(n 1)/2. L 2 ( B r) This is equivalent to where w X 0 r p X 0 κ 2 (2ε) 1/2 (29) L 2 ( B 1) w X 0 r (X) = u (rx + X 0) r κ. 41

42 Covering K with a finite number of balls B δε(x 0) (X 0) for some X 0 K, we obtain that (29) holds for all X 0 K with r rε K. We claim that, if X 0 K and 0 < r < rε K then w X 0 r p X 0 C ε with C ε 0 as ε 0. (30) L (B 1/2) κ To prove the claim, observe that the two functions w X 0 r and p X 0 κ are both solution of our zero thin obstacle problem, uniformly bounded in C 1,α (B ± 1 ). If (30) were not true, by compactness, we can construct a sequence of solutions converging to a non trivial zero trace solutions (from (29)). The uniqueness of the solution of the thin obstacle problem with Dirichlet data implies a contradiction. It is easy to check that the claim implies the second part of the lemma. We are now in position to prove Theorem 36. The proof uses the Whitney s extension theorem (see [66] or [67]) and the implicit function theorem. We recall that the extension theorem prescribes the compatibility conditions under which there exists a C k function f in R N having prescribed derivatives up to the order k on a given closed set. Since our reference set is Σ κ (u), we first need to show that Σ κ (u) is a countable union of closed sets (an F σ set). This is done in the next Lemma. Lemma 42 (Topological structure of Σ κ (u)) Σ κ (u) is a F σ set. Proof. Let E j be the set of points X 0 Σ κ (u) B 1 1/j such that 1 j ρκ sup u (X) < jρ κ (31) X X 0 =ρ for 0 < ρ < 1 X 0. By non degeneracy and (27) we know that Σ κ (u) j 1 E j. We want to show that E j is a closed set. Indeed, if X 0 E j then X 0 satisfies (31), and we only need to show that X 0 Σ κ (u), i.e., from Theorem 2.9.1, that Φ X0 (0+; u) = κ. Since the function X Φ X (0+; u) is upper-semicontinuous we deduce that Φ X0 (0+; u) = κ κ. If we had κ > κ, we would have u (X) X X 0 κ in B 1 X0 (X 0 ), which contradicts the estimate from below in (31). Σ κ (u). We are now in position for the proof of Theorem 36. Thus κ = κ and X 0 Proof of Theorem 36. We divide the proof into two steps. Recall that Σ κ (u) = F κ (u) if κ = 2m. 42

43 Step 1. Whitney s extension. For simplicity it is better to make a slight change of notations, letting y = x n+1 and X = (x 1,..., x n, x n+1 ). Let K = E j be one of the compact subsets of Σ κ (u) constructed in Lemma 42. We can write p X0 κ (X) = α =κ a α (X 0 ) X α. α! The coefficients a α (X) are continuous on Σ κ (u) by Theorem 35. Since u = 0 on Σ κ (u) we have p X 0 κ (X X 0 ) σ ( X X0 ) X X 0 κ X K. For every multi-index α, 0 α κ, define: { 0 if 0 < α < κ f α (X) = a α (X) if α = κ X Σ κ (u). We want to construct a function f C κ ( R n+1), whose derivatives α f up to the order κ are prescribed and equal to f α on K. The Whitney extension theorem states that this is possible if, for all X, X 0 K, the following coherence conditions hold for every multi-index α, 0 α κ : with f α (X) = β κ α f α+β (X 0 ) (X X 0 ) β + R α (X, X 0 ) (32) β! R α (X, X 0 ) σ K α ( X X 0 ) X X 0 κ α, (33) where σ K α is a modulus of continuity. Claim: (32) and (33) hold in our case. Proof. Case α = κ. Then we have R α (X, X 0 ) = a α (X) a α (X 0 ) and therefore R α (X, X 0 ) σ α ( X X 0 ) by the continuity on K of the map X p X κ. Case 0 α < κ. We have R 0 (X, X 0 ) = γ>α, γ =κ a γ (X 0 ) (γ α)! (X X 0) γ α = α p X0 κ (X X 0 ). Now, suppose that there exists no modulus of continuity σ α such that (33) holds for all X, X 0 K. Then, there is δ > 0 and sequences X i, X0 i K with X i X0 i = ρ i 0 and such that α p X0 κ (X X 0 ) δ X i X0 i κ α. (34) 43

44 Consider the rescalings w i (X) = u ( X i 0 + ρ i X ) ρ κ i, ξ i = Xi X i 0 ρ i. We may assume that X0 i X 0 K and ξ i ξ 0 B 1. From Lemma 41 we have that w i (X) p Xi 0 κ (X) σ (ρ i X ) X κ and therefore w i (X) converges to p Xi 0 κ (X), uniformly in every compact subset of R n+1. Note that, since X i, X0 i K, the inequalities (31) are satisfied there. Moreover, we also have that similar inequalities are satisfied for the rescaled function w i at 0 and ξ i. Thus, passing to the limit, we deduce that 1 j ρκ sup p X 0 κ (X) < jρ κ, 0 < ρ < +. X X 0 =ρ This implies that ξ 0 is a point of frequency κ = 2m for the polynomial p X0 κ that, from Theorem 35, we infer that ξ 0 Σ κ (p ξ 0 κ ). In particular, so α p ξ 0 κ = 0 for α < κ. However, dividing both sides of (34) by ρ κ α i obtain α p ξ 0 κ δ, a contradiction. This ends the proof of the claim. and passing to the limit, we Step 2. Implicit function theorem. Applying Whitney s Theorem we deduce the existence of a function f C κ ( R n+1) such that α f = f α on E j for every α κ. Suppose now X 0 Σ d κ (u). This means that d = dim { ξ R n : ξ x p X0 κ (x, 0) 0 }. Then there are n d linearly independent unit vectors ν i R n, such that ν i x p X0 κ (x, 0) is not identically zero. This implies that there exist multi-indices β i of order β i = κ 1 such that This can be written as νi ( βi p X0 κ (X 0 )) 0. νi ( βi f (X 0 )) 0, i = 1,..., n d. (35) 44

45 On the other hand, we have } Σ d κ (u) E j i=1,...,n d { βi f = 0. From (35) and the implicit function theorem, we deduce that Σ d κ (u) E j is contained in a d dimensional C 1 manifold in a neighborhood of X 0. Since Σ κ (u) = E j we conclude the proof Non zero obstacle The above differentiability and the structure theorems can be extended to the non zero obstacle case, if ϕ C k,1 (B 1) for some integer k 2 ([36]). A crucial tool is once more a frequency formula which generalizes the one in Theorem 16. First, we introduce a useful change of variable to reduce to the case in which the Laplacian is very small outside the coincidence set. Let u be a solution of the obstacle problem and set: v (x, y) = u (x, y) Q k (x, y) (ϕ (x) Q k (x)), where Q k is the k-th Taylor polynomial of ϕ at the origin and Q k is its even harmonic extension to all of R n+1. Now v can be evenly extended to y < 0 and then v M x k u y H n in D (B B 1 1 ). The generalized frequency formula takes the following form: If v ( ) C 1 B + 1 M/2, then there exists r M and C M such that the function r Φ k (r; u) = ( r + C M r 2) d dr log max { K (r), r n+2k} is nondecreasing for 0 < r < r M. Also the Weiss and Monneau monotonicity formulas have to be modified accordingly to take into account the perturbation introduced by the non-zero obstacle. Indeed we have: For κ k, 0 < r < r M, C M > 0, d dr W κ (r; u) C M and for κ = 2m < k, 0 < r < r m, C M > 0, d ( ) dr M κ (r; u, p κ ) C M 1 + p κ L 2 (B 1). Coherently, the Differentiability Theorem 35 and the Structure Theorem 36 hold for κ = 2m < k. This limitation is due to the fact that for points in F κ (u) even the blow-ups are not properly defined. 45

46 Thus, the analysis of the free boundary around a singular point is rather satisfactory. It remains open the study of the nonregular, nonsingular points, i.e. the set Ξ (u) = {F κ (u) : κ > 32 }, κ 2m, m 1, integer. It should be noted that Ξ (u) could in principle be a large part of F (u). Another important issue that remains open is whether F (u) has Hausdorff dimension n 1. Further analysis of F (u) clearly depends on the possible values that the frequency κ may attain. A partial classification of convex global solutions excludes the interval (3/2, 2) from the range of possible values of κ. As observed in ([36]) it is plausible that the only possible values are κ = 2m 1 2 or κ = 2m, m 1, integer. This is indeed true in dimension 2 (n = 1), see remark in [36] A global regularity result (fractional Laplacian) In ([12]), Barrios, Figalli and Ros-Oton consider both the global and the local version of the Signorini problem for the operator L a, thus covering also the case of the obstacle problem for the fractional Laplacian. We limit ourselves to briefly describe their main results and ideas in the case of the local Signorini problem. Under basically two main assumptions the authors can give a complete picture of the free boundary, recovering a result completely analogous to the classical case (s = 1). The first assumption is a strict concavity of the obstacle, the same assumption needed in the case of the classical obstacle problem. The second one prescribes zero boundary values of the solution and it turns out to be a crucial assumption. Precisely their main result in the case of the Signorini problem is the following. As in Section 2.7 we define the blow-ups of v at X 0 by vr X0 (X) = v (r(x X 1/2 0)) d r = (r n a u 2 y dσ) a. (36) d r B r(x 0) Theorem 43 Let ϕ : B 1 R with ϕ B < 0, and u : B 1 R n+1 R be a 1 solution of the following problem: u (x, 0) ϕ (x) on B 1 L a u = 0 in B 1 \ ({y = 0} {u = ϕ}) L a u 0 in B 1 u = 0 on B 46

47 with u (x, y) = u (x, y). Assume that ϕ C 3,γ (B 1), ϕ c 0 < 0 in {ϕ > 0}, {ϕ > 0} B 1 (37) for some c 0 > 0 and γ > 0. Then, at every singular point the blow-up of u is a homogeneous polynomial of degree 2, and the free boundary can be decomposed as F (u) = F 1+s (u) F 2 (u) where F 1+s (u) (resp. F 2 (u)) is an open (resp. closed) subset of F (u). Moreover, F 1+s (u) is a (n 1)-dimensional manifold of class C 1,α, while F 2 (u) can be stratified as the union of sets { F k 2 (u) } k=0,1,...,n 1, where F k 2 (u) is contained in a k-dimensional manifold of class C 1. The regularity of F 1+s (u) comes from [22]. Here the main points are that the nonregular points are only singular points, that the blow-up at these points has homogeneity 2, that F 2 (u) can be stratified into C 1 manifolds and that each F k 2 (u) is completely contained in a k-dimensional C 1 manifold. The key lemma is the following nondegeneracy result. It says that u detaches quadratically from the obstacle putting out of game all possible frequencies greater than 2. Lemma 44 Let u be as in Theorem 43. Then there exists constants c 1, r 1 > 0 such that the following holds: for any X 0 F (u) we have sup (u (x, 0) ϕ (x)) c 1 r 2, 0 < r < r 1. x B r (x0) Proof. Since u > 0 on the contact set, compactly contained in B 1, we deduce that ϕ h 0 > 0 on {u = ϕ}. For r 1 > 0 small, from the properties of ϕ, in the set U 1 = {x : u (x, 0) > ϕ (x), dist (x, F (u)) 2r 1 } B 1, we have ϕ > 0. Take x 1 U 1, with dist(x, F (u)) r 1 and consider the barrier function given by w (x, y) = u (x, y) ϕ (x) c 0 ( x x1 + y 2) 2n + 2 (1 + a) where c 0 is as in (37). Note that w (x 1, 0) > 0 and w < 0 on {u = ϕ} {y = 0}. We want to apply the maximum principle in the set U = B r (x 1, 0) \ ({u = ϕ} {y = 0}) B 1. Since L a u = 0 and ϕ (x) c 0, we have L a w (x, y) = L a u (x, y) y a ( ϕ (x) + c 0 ) 0. 47

48 Noticing that U = B r (x 1, 0) {u = ϕ} {y = 0}, by the maximum principle we infer that Letting x 1 x 0 we get 0 < w (x 1, 0) sup U = sup (u ϕ) B r(x 1,0) w = sup w = U sup w B r(x 1,0) c 0 2n + 2 (1 + a) r2. c 0 sup (u ϕ) sup (u ϕ) B r(x 0,0) B r(x 0,0) 2n + 2 (1 + a) r2. (38) To conclude the proof we must show that the above supremum is attained on y = 0. To prove it we show that u y 0 in B 1 +. Here comes into play the zero boundary condition. Since L a u 0 and L a u = 0 outside the contact set {u = ϕ}, by the maximum principle it follows that u 0 in B 1 and u attains its maximum on {u = ϕ}. Using again that u = 0 on B 1 and that u (x, y) = u (x, y), we deduce that Moreover y a u y 0 and lim y 0 + y a u y (x, y) 0 on {u = ϕ}. lim y a u y (x, y) = 0 on {u > ϕ}. y 0 + Now, by direct computation, one can check that L a (y a u y ) = 0 in B 1 +. By the maximum principle we infer that y a u y 0 in B 1 +. Thus, u (x, y) is decreasing with respect to y in B 1 +. Since u is even in y, this yields and (38) finally gives sup (u (x, 0) ϕ (x)) = x B r (x0) u (x, y) u (x, 0) in B 1 sup (u ϕ) B r(x 0,0) c 0 2n + 2 (1 + a) r2. Having the above nondegeneracy at hands, to proceed further once again the main tools are frequency and monotonicity formulas, adapted to take into account the homogeneity of the solution. For a free boundary point x 0 F (u), the change of variable v x0 (x, y) = u (x, y) ϕ (x) + 1 { ϕ (x0 ) y 2 + ϕ (x 0 ) (x x 0 ) y 2} 2 (1 + a) reduces to the case L a v x0 C y a x x 0 1+γ (39) 48

49 outside the coincidence set and takes care of the errors in the frequency/monotonicity formulas, due to the presence of a non-zero obstacle. Note that v x0 (x 0, 0) = u (x, 0) ϕ (x) and that v x0 depends continuously on x 0 since ϕ C 3,γ. The non-degeneracy of u translates into sup v x0 (x, 0) cr 2. B r (x0) Moreover, exploiting (39) and a weak Harnack inequality (see [29]) one also gets, for small r, y a v x0 (x, y) 2 dxdy c 1 r n+a+5. (40) B r(x 0,0) Setting x 0 = 0, v = v 0 and K (r) = the function B r(0,0) y 2 v 2, r Φ (r; v) = ( r + C 0 r 2) d dr log max { K (r), r n+a+4+2γ} is monotone non-decreasing for small r, and a suitable constant C 0 > 0. Let Φ (0+; v) = n + a + 2m. Now, the above non-degeneracy estimates implies that either m = 1 + s or m = 2. It is clearly enough to show that m 2. Indeed, the monotonicity of Φ gives n + a + 2m Φ (r; v) and integrating we get K (r) Cr n+a+2m. A further integration gives y a v x0 (x, y) 2 dxdy c 1 r n+a+2m+1 B r(0,0) that together with (40) implies m 2. At this point, following the strategy in [22], it is possible to show that, up to a subsequence, the blow-ups v r in (36) converge as r 0 + to a function, which is nonnegative on y = 0 and homogeneous of degree m. Moreover: a) either there exist c > 0 and a unit vector ν such that m = 1 + s and v 0 (x, 0) = c (x ν) 1+s + b) or m = 2 and v 0 (x, 0) is a polynomial of degree 2 49

50 and the origin is a singular point. Uniqueness of the blow-ups can be proved by a suitable modification of the Weiss and Monneau monotonicity formulas. As consequence, there exists a modulus of continuity ω : R + R + such that, for all x 0 belonging to the singular set F 2 (u) of the free boundary, we have u (x) ϕ (x) = p x0 2 (x x 0) + ω ( x x 0 ) x x 2 0 (41) for some polynomial p x0 2 (x) = (Ax0 x, x), A R n n, symmetric and nonnegative, A 0. In addition, the mapping F 2 (u) x 0 p x0 2 is continuous with y a (p x1 2 px0 2 ) ω ( x 1 x 0 ) x 1, x 0 F 2 (u). B 1 Concerning the regularity of F 2 (u), given any point x 0 in this set, from (41), the blow-up of u ϕ coincides with the polynomial p x0 2 (x) = x, x). We (Ax0 stratify F 2 (u) according to the dimension of ker A : F k 2 (u) = {x 0 F 2 (u) : dim ker A x0 = k} k = 0, 1,..., n 1. Then, reasoning as in the classical obstacle case (see [18]), one can show that for any x 0 F k 2 (u) there exists r = r x0 > 0 such that F k 2 (u) B r (x 0 ) is contained in a connected k dimensional C 1 manifold. This concludes the proof of Theorem Comments and further reading In recent time and while we were writing this survey several important results have been established. We briefly mention some of the more strictly related to our presentation. Different approaches to the regularity of the free boundary. When the thin manifold is non-flat, by a standard procedure one is lead to a Signorini problem, but for a divergence form operator with variable coefficient matrix A(x) = [a ij (x)]. For instance, when M satisfies the minimal smoothness assumption C 1,1 = W 2,, then A(x) is C 0,1 = W 1,, i.e., Lipschitz continuous. The approach described above in the case A(x) I n to establish the C 1,α regularity of the regular free boundary, based on differentiating the equation for the solution v in tangential directions e R n 1 and establishing directional monotonicity, does not work well for variable coefficients, particularly when the obstacle ϕ 0. For coefficients A(x) W 1,, and the obstacle ϕ W 2,, the optimal C 1,1/2 loc (Ω ± M) regularity of the solutions was established by Garofalo and Smit Vega Garcia in [39] by means of monotonicity formulas of Almgren s type. The C 1,α smoothness of the regular free boundary has been obtained by Garofalo, Petrosyan and Smit Vega Garcia in [37]. The two central tools are a Weiss type monotonicity formula and an epiperimetric inequality, which allow to control the homogeneous blow-ups. The latter results, in the special 50

51 case A(x) I n, have also been established by Focardi and Spadaro ([31]). A different approach, based on Carleman estimates, to the optimal regularity of the solutions and C 1,α regularity of the free boundary for A(x) W 1,p, p > 2n, and vanishing obstacle is used by Koch, Rüland and Shi in [44] and [45] (more recently the authors were able to extend these results to non-zero obstacles in W 2,p, p > 2n). Higher regularity. Real analyticity of the regular part of the free boundary for the thin obstacle (Signorini) problem has been proved by Koch, Petrosyan and Shi in [43] via a partial hodograph-legendre transformation, subsequently extended by Koch, Rüland and Shi to the fractional Laplacian operator in ([46]). The lack of regularity of this map can be overcome by providing a precise asymptotic behavior at a regular free boundary point. The Legendre transforms (on which one reads the regularity of the free boundary) satisfies a subelliptic equation of Baouendi-Grushin type and its analyticity is achieved by using the L p theory available for this kind of operator. A different approach to higher regularity for the thin obstacle and also to one-phase free boundary problems is due to De Silva and Savin in [27] and is based on a higher order Boundary Harnack principle. Jhaveri and Neumayer in [41] extend this approach to the fractional Laplacian obstacle problem. The higher regularity of the free boundary in the variable coefficient case has been established in [47]. More general problems and operators. Allen, Lindgren and Petrosyan in [1] consider the two phase problem for the fractional Laplacian and prove optimal regularity of the solution and separation of the positive and negative phase. Operators with drift in the subcritical regime s (0, 1/2) are considered by Petrosyan and Pop in [54] where optimal regularity of the solution is proved. The regularity of the free boundary is addressed by Garofalo et al. in [38]. We emphasize that the presence of the drift the extension operator exhibits some singularities and makes the problem quite delicate. The results in [22] have been extended to obstacle problem for integrodifferential operators by Caffarelli, Ros-Oton and Serra in [20]. Here the authors develop an entirely non local powerful approach, independent of any monotonicity formulas. Korvenpää, Kuusi and Palatucci in [48] consider the obstacle problem for a class of nonlinear integro-differential operators including the fractional p Laplacian. The solution exists, it is unique and inherits regularity properties (such as Hölder continuity) from the obstacle. 4 Parabolic obstacle problems In this section we will focus on time-dependent models, which can be thought of as parabolic counterparts of the systems (1) and (2). We emphasize that, although their time-independent versions are locally equivalent (for s = 1/2), the problems we are about to describe are very different from each other. 51

52 4.1 The parabolic fractional obstacle problem As mentioned above, one of the motivations behind the recent increased interest in studying constrained variational problems with a fractional diffusion comes from mathematical finance. Jump-diffusion processes allow, in fact, to take into account large price changes, and appear to be better suited to model market fluctuations. An American option gives its holder the right to buy a stock or basket of stocks at a given price prior to, but not later than, a given time T > 0 from the time of inception of the contract. If v(x, t) denotes the price of an American option with a payoff ψ at time T, then v will be a viscosity solution to the obstacle problem { min{lv, v ψ} = 0 (42) v(t ) = ψ. Here L is a backward parabolic integro-differential operator of the form Lv = v t rv b v + ( ) s v + Hv, s (0, 1), where r > 0, b R n, and H is a non-local operator of lower order with respect to ( ) s. This problem was studied by Caffarelli and Figalli in the paper [19], where regularity properties are established for the model equation { min{ v t + ( ) s v, v ψ} = 0 on [0, T ] R n v(t ) = ψ on R n (43). The advantage of considering (43) is twofold. On the one hand, the absence of the transport term allows to prove that solutions have the same regularity as in the stationary case for all values of s (0, 1), even if, when s (0, 1/2), the time derivative is of higher order with respect to the elliptic term ( ) s v. In addition, it is feasible that the regularity theory established for the model equation (43) could be adapted to the solutions to (42) when s > 1/2. It should be noted that when s < 1/2, the leading term becomes b v, and there is no expectation of a regularity theory. In order to proceed, we need to introduce the relevant function spaces. Definition 3 Given α, β (0, 1) and [a, b] R, we say that w C α,β t,x ([a, b] R n ) if w C α,β t,x ([a,b] Rn ) := w L ([a,b] R n ) + [w] C α,β t,x ([a,b] Rn ) w loglip t C β x ([a, b] R n ) if = w L ([a,b] R n ) + sup [a,b] R n w(t, x) w(t, x ) t t α + x x β <. w loglipt Cx β ([a,b] R n ) := w w(t, x) w(t, x ) L ([a,b] R n )+ sup [a,b] R n t t ( 1 + log t t ) + x x <. β 52

53 We will also say that w C α 0+,β t,x ([a, b] R n ) if w C α ε,β t,x ([a, b] R n ) for all ε > 0; w C α,β t,x ((a, b] R n ) if w C α,β t,x ([a + ε, b] R n ) for all ε > 0. In what follows, ψ : R n R + will be a globally Lipschitz function of class C 2 satisfying ψ R n (1+ x ) < and ( ) s ψ L (R n ). For s (0, 1) we let n+2s u : [0, T ] R n R be a continuous viscosity solution to the obstacle problem { min{u t + ( ) s u, u ψ} = 0 on [0, T ] R n u(0) = ψ on R n (44). Existence and uniqueness of solutions can be shown either by probabilistic techniques, or by approximating the equation via a penalization method. We are now ready to state the main result from [19]. Theorem 45 Assume that ψ C 2 (R n ) satisfies ψ L (R n ) + D 2 ψ L (R n ) + ( ) s ψ C 1 s x (R n ) <, and that u solves (44). Then u is globally Lipschitz in space-time on [0, T ] R n, and satisfies C 1 s x u t loglip t C 1 s x ((0, T ] R n ), ( ) s u loglip t C 1 s x ((0, T ] R n ) if s 1/3, u t C 1 s 2s 0+ t,x ((0, T ] R n ), ( ) s u C 1 s 2s t,x ((0, T ] R n ) if s > 1/3. Some remarks are in order. First of all, comparison with Corollary 25 shows that, at least in terms of spacial regularity, this result is optimal. Secondly, the criticality of s = 1/3 is a consequence of the invariance of the operator t + ( ) s under the scaling (t, x) (λ 2s t, λx). Hence, the spacial regularity naturally corresponds to time regularity C 1 s 2s t when 1 s < 1, or s > 1/3. In addition, the time regularity is almost optimal in the case s = 1/2 (as it is possible to construct traveling waves which are C 1+1/2 both in space and time), as well as in the limit s 1 (since, when s = 1, it is well known that solutions are C 1,1 in space and C 1 in time). Several basic properties of the solution u, such as global regularity in spacetime, semi-convexity in space, and the boundedness of ( ) s u, follow from a comparison principle, which in turn is established using a penalization method. Combining the semi-convexity of u(t) with the L bound on ( ) s u(t), it is possible then to deduce the C 1 regularity in time of solutions when s 1/2. The next step toward the proof of Theorem 45 is to prove a C α+2s -regularity result in space, which, roughly speaking, says the following. Let v : R n R be a semiconvex function touching from above an obstacle ψ : R n R of class C 2. 2s 53

54 If ( ) s v is non-positive outside the contact set and non-negative on it, then v detaches from ψ in a C α+2s fashion, with α > 0 depending exclusively on s. More precisely, assume that v, ψ : R n R are two globally Lipschitz functions with v ψ satisfying the following conditions: A1. D 2 ψ L (R n ) := C 0 < ; A2. ( ) s ψ C 1 s x (R n ) < ; A3. v ψ, ( ) s v L (R n ); A4. The function v + C 0 x 2 /2 is convex (we say that v is C 0 -semiconvex); A5. v is smooth and ( ) s v 0 inside the open set {v > ψ}; A6. ( ) s ψ L (R n ) ( ) s v 0 on {v = ψ}. One has then the following: Theorem 46 There exist C > 0 and α (0, 1), depending only on C 0, ( ) s ψ C 1 s x (R n ), v ψ L (R n ), and ( ) s v L (R n ), such that sup v ψ Cr α+2s, sup ( ) s vχ {v=ψ} Cr α (45) B r(x) B r(x) for all r 1 and for every x {u = ψ}. The strategy of the proof is analogous to the one used in [7]. It is based on growth estimates for the L a -harmonic extension of v, i.e. the solution to (4), satisfying (with slight abuse of notation) v(x, 0) = v(x), with v(x) as above. Having shown that ( ) s vχ {v=ψ} grows at most as r α near any free boundary point, it is not difficult to show that ( ) s vχ {v=ψ} C α x (R n ): Corollary 47 There exist C > 0 and α (0, 1 s], depending only on C 0, ( ) s ψ C 1 s x (R n ), v ψ L (R n ), and ( ) s v L (R n ), such that ( ) s vχ {v=ψ} C α x (R n ) C. At this point we consider the function w : R n R + R, which solves the Dirichlet problem { L a w = 0 on R n R +, w(x, 0) = ( ) s v(x)χ {v=ψ} (x) on R n. Since w(x, 0) 0, the maximum principle implies w 0 everywhere. Assume now that 0 is a free boundary point. Given that ( ) s v(x) is globally bounded by (A3) above, it follows from the Poisson representation formula for w (see [23]) and Corollary 47 that sup w(x, y) Cr α, x 2 +y 2 r 2 for some uniform constant C and some α (0, 1 s]. establish the following sharp growth estimate: Our next goal is to 54

55 Proposition 48 There exists C > 0 depending only on C 0, ( ) s ψ C 1 s x (R n ), v ψ L (R n ), and ( ) s v L (R n ), such that sup w(x, y) Cr 1 s. x 2 +y 2 r 2 A crucial ingredient in the proof of Proposition 48 is the following monotonicity formula (compare with Lemma 10). Lemma 49 Let w be as above and define φ (r) = 1 r 2(1 s) B r + y a w (X) 2 dx. X n a 1 Then there exists a constant C, depending only on C 0, ( ) s ψ C 1 s x (R n ), v ψ L (R n ), and ( ) s v L (R n ), such that for all r 1. Here, δ α = 1 4 φ(r) C [1 + r 2α+δα a 1 ] ( ). α α+2s α 2 Proof of Proposition 48. Using an approximation argument, it is possible to show that the function w(x, y) = ( w(x, y ) r α+δα) ( nc ) 0 y 1+a 1 + α is globally L a -subharmonic. Moreover, because of Lemma 49, it vanishes on more than half of the n-dimensional disc B r {0}. One can then apply the weighted Poincaré inequality established in [29] to show ( w) 2 y a dx Cr n+2 [φ(r) + 1] B + r for all r 1. Combining this estimate with the L a -subharmonicity of w, and Lemma 49 again, we infer sup( w) 2 C ( w) 2 y a dx C[r 1+α + r 2α+δα ]. B + r/2 r n+1 a B + r But 1 + α = 2(1 s), and therefore sup w C sup w + r α+δα + r 1+a C[r 1 s + r α+δα/2 ]. B + r/2 B + r/2 Arguments similar to the ones in the proof of Corollary 47 yield w C β α x (R n ) C, 55

56 with β α = min{α + δ α /2, 1 s}. An iteration gives the desired conclusion. Arguing as in the proof of Corollary 47, we obtain ( ) s vχ {v=ψ} C 1 s x (R n ) C. (46) Next, we apply (46) to v = u(t), obtaining the following result. Proposition 50 Let u be a solution to (44), with ψ C 2 (R n ) satisfying (A1) and (A2) above. Then there exists a constant C T > 0, depending on T, D 2 ψ L (R n ), and ( ) s ψ C 1 s x (R n ), such that sup ( ) s u(t)χ {u(t)=ψ} C 1 s x (R n ) C T. t [0,T ] Finally, to conclude the proof of Theorem 45 one exploits the fact that u is a solution of the parabolic equation u t + ( ) s u = ( ( ) s u ) χ {u=ψ} on (0, T ] R n. (47) Proposition 50 ensures that the right-hand side of (47) is in L ( (0, T ]; C 1 s x (R n ) ). By parabolic regularity theory we infer u t, ( ) s u L ( (0, T ]; C 1 s 0+ x (R n ) ). The desired Hölder regularity in time follows from a bootstrap argument which uses equation (47) again. 4.2 The parabolic Signorini problem In this section we will give an overview of the time-dependent analogue of (2), i.e., the parabolic Signorini problem, following the ideas in [26] Statement of the problem Given a domain Ω in R n, n 2, with a sufficiently regular boundary Ω, let M be a relatively open subset of Ω (in its relative topology), and set S = Ω\M. We consider the solution of the problem v t v = 0 in Ω T := Ω (0, T ], (48) v φ, ν v 0, (v φ) ν v = 0 on M T := M (0, T ], (49) v = g on S T := S (0, T ], (50) v(, 0) = φ 0 on Ω 0 := Ω {0}, (51) where ν is the outer normal derivative on Ω, and φ : M T R, φ 0 : Ω 0 R, and g : S T R are prescribed functions satisfying the compatibility conditions: φ 0 φ on M {0}, g φ on S (0, T ], and g = φ on S {0}, see Fig. 2. Classical examples where Signorini-type boundary conditions appear are 56

57 Ω T M T v = φ νv 0 v > φ νv = 0 Γ(v) v = g v = φ 0 Figure 2: The parabolic Signorini problem the problems with unilateral constraints in elastostatics (including the original Signorini problem [61, 30]), problems with semipermeable membranes in fluid mechanics (including the phenomenon of osmosis and osmotic pressure in biochemistry), and the problems on the temperature control on the boundary in thermics. We refer to the book of Duvaut and Lions [28] for further details. Another historical importance of the parabolic Signorini problem is that it serves as one of the prototypical examples of evolutionary variational inequalities. We thus say that a function v W 1,0 2 (Ω T ) solves (48) (51) if v (w v) + t v(w v) 0 for every w K, Ω T v K, t v L 2 (Ω T ), v(, 0) = φ 0, where K = {w W 1,0 2 (Ω T ) w φ on M T, w = g on S T } (please see below for the definitions of the relevant parabolic functional classes). The existence and uniqueness of such v, under some natural assumptions on φ, φ 0, and g can be found in [14, 28, 3, 4]. Similarly to the elliptic case, we are interested in: the regularity properties of v; the structure and regularity of the free boundary Γ(v) = MT {(x, t) M T v(x, t) > φ(x, t)}, where MT indicates the boundary in the relative topology of M T. Concerning the regularity of v, it has long been known that the spatial derivatives xi v, i = 1,..., n, are α-hölder continuous on compact subsets of Ω T M T, for some unspecified α (0, 1). In the parabolic case, this was first proved by Athanasopoulos [5], and subsequently by Uraltseva [64] (see also [3]), 57

58 under certain regularity assumptions on the boundary data, which were further relaxed by Arkhipova and Uraltseva [4]. One of the main objectives of this section is to establish, in the parabolic Signorini problem, and for a flat thin manifold M, that v H 3/2,3/4 loc (Ω T M T ), see Theorem 56 below. The proof, which we only sketch here, is inspired by the works [10] and [22] described in Section 2. For further details, we refer to [26]. Before proceeding, we introduce the relevant parabolic function spaces and notations. We use notations similar to those in the classical book of Ladyzhenskaya, Solonnikov, and Uraltseva [49]. The class C(Ω T ) = C 0,0 (Ω T ) is the class of functions continuous in Ω T with respect to parabolic (or Euclidean) distance. Further, given for m Z + we say u C 2m,m (Ω T ) if for α + 2j 2m x α j t u C 0,0 (Ω T ), and define the norm u C 2m,m (Ω T ) = sup x α j t u(x, y). (x,t) Ω T α +2j 2m The parabolic Hölder classes H l,l/2 (Ω T ), for l = m + γ, m Z +, 0 < γ 1 are defined as follows. First, we let u (0) Ω T = u (0) Ω T = sup (x,t) Ω T u(x, t), u (m) Ω T = α +2j=m α x j t u (0) Ω T, u (β) u(x, t) u(y, t) x,ω T = sup (x,t),(y,t) Ω T x y β, 0 < β 1, 0< x y δ 0 u (β) u(x, t) u(x, s) t,ω T = sup (x,t),(x,s) Ω T t s β, 0 < β 1, 0< t s <δ 2 0 u (l) x,ω T = x α j t u (γ) x,ω T, u (l/2) t,ω T = u (l) Ω T α +2j=m m 1 α +2j m = u (l) x,ω T + u (l/2) t,ω T. α x j t u ((l α 2j)/2) t,ω T, Then, we define H l,l/2 (Ω T ) as the space of functions u for which the following norm is finite: m u H l,l/2 (Ω T ) = u (k) + u (l) Ω T. k=0 The parabolic Lebesgue space L q (Ω T ) indicates the Banach space of those measurable functions on Ω T for which the norm ( ) 1/q u Lq(Ω T ) = u(x, t) q dxdt Ω T 58 Ω T

59 is finite. The parabolic Sobolev spaces Wq 2m,m (Ω T ), m Z +, denote the spaces of those functions in L q (Ω T ), whose distributional derivative x α j t u belongs to L q (Ω T ), for α + 2j 2m. For x, x 0 R n, t 0 R we let x = (x 1, x 2,..., x n 1 ), x = (x 1, x 2,..., x n 2 ), x = (x, x n ), x = (x, x n 1 ), and B r (x 0 ) = {x R n x < r} B r ± (x 0 ) = B r (x 0 ) R n ± B r(x 0 ) = B r (x) R n 1 Q r (x 0, t 0 ) = B r (x 0 ) (t 0 r 2, t 0 ] (Euclidean ball) (Euclidean halfball) ( thin ball) (parabolic cylinder) Q r(x 0, t 0 ) = B r(x 0 ) (t 0 r 2, t 0 ] ( thin parabolic cylinder) Q ± r (x 0, t 0 ) = B ± r (x 0 ) (t 0 r 2, t 0 ] Q r (x 0, t 0 ) = B r (x 0 ) (t 0 r 2, t 0 ] S r = R n ( r 2, 0] (parabolic halfcylinders) (parabolic strip) S ± r = R n ± ( r 2, 0] (parabolic halfstrip) S r = R n 1 ( r 2, 0] ( thin parabolic strip) When x 0 = 0 and t 0 = 0, we omit the centers x 0 and (x 0, t 0 ) in the above notations. Since we are mostly interested in local properties of the solution v of the parabolic Signorini problem and of its free boundary, we focus our attention on solutions in parabolic (half-)cylinders. Definition 4 Given φ H 2,1 (Q 1), we say that v S φ (Q + 2,1 1 ) if v W2 (Q + 1 ) L (Q + 1 ), v Hα,α/2 (Q + 1 Q 1) for some 0 < α < 1, and v satisfies and v t v = 0 in Q + 1, (52) v φ 0, xn v 0, (v φ) xn v = 0 on Q 1, (53) (0, 0) Γ (v) := Q 1 {(x, t) Q 1 v(x, 0, t) = φ(x, t), xn v(x, 0, t) = 0}, (54) where Q 1 is the boundary in the relative topology of Q 1. Our very first step is the reduction to vanishing obstacle. The difference v(x, t) φ(x, t) satisfies the Signorini conditions on Q 1 with zero obstacle, but at an expense of solving a nonhomogeneous heat equation instead of the homogeneous one. This difference may then be extended to the strip S 1 + by multiplying it by a suitable cutoff function ψ. The resulting function will satisfy u t u = f(x, t) in S + 1, 59

60 with f(x, t) = ψ(x)[ φ t φ] + [v(x, t) φ(x, t)] ψ + 2 v ψ. (55) Remark 4 It is important to observe that, for smooth enough φ, the function f is bounded in S + 1. With this process, we arrive at the following notion of solution. Definition 5 A function u is in the class S f (S 1 + ), for f L (S 1 + ), if u W 2,1 2 (S 1 + ), u Hα,α/2 (S 1 + S 1), u has compact support and solves u t u = f in S + 1, (56) u 0, xn u 0, u xn u = 0 on S 1, (57) (0, 0) Γ(u) = {(x, t) S 1 : u(x, 0, t) > 0}. (58) Monotonicity of the generalized frequency function Similarly to the elliptic case, one of the central results towards the study of regularity properties (both of the solution and of the free boundary) is a generalization of Almgren s frequency formula, see Theorem 51 below. As it is well known, the parabolic counterpart of Almgren s formula was established by Poon [58], for functions which are caloric in an infinite strip S ρ = R n ( ρ 2, 0]. Poon s parabolic frequency function is given by N u (r) = i u( r 2 ) h u ( r 2 ), (59) where h u (t)= u(x, t) 2 G(x, t)dx, R n + i u (t)= t R n + u(x, t) 2 G(x, t)dx, for any function u in the parabolic half-strip S 1 + for which the integrals involved are finite. Here G denotes the backward heat kernel on R n R {( 4πt) G(x, t) = n 2 e x2 4t, t < 0, 0, t 0. We explicitly remark that Poon s monotonicity formula cannot be directly applied in the present context, since functions in the class S f (S 1 + ) (see Definition 5) are not caloric functions. It should also be noted that the time dependent case of the Signorini problem presents substantial novel challenges with respect to the stationary setting. These are mainly due to the lack of regularity of the solution in the t-variable, a fact which makes the justification of differentiation formulas and the control of error terms quite difficult. To overcome 60

61 these obstructions, (Steklov-type) averaged versions of the quantities involved are introduced in the main monotonicity formulas. This basic idea allows to successfully control the error terms. More precisely, we introduce the quantities H u (r)= 1 0 r 2 h u (t)dt = 1 r r 2 2 I u (r)= 1 0 r 2 r 2 i u (t)dt = 1 r 2 S + r S + r u(x, t) 2 G(x, t)dxdt, t u(x, t) 2 G(x, t)dxdt, One further obstruction is represented by the fact that the above integrals may become unbounded near the endpoint t = 0, where G becomes singular. To remedy this problem we introduce truncated versions of H u and I u : Hu(r) δ = 1 δ 2 r 2 r 2 Iu(r) δ = 1 δ 2 r 2 r 2 r 2 h u (t)dt = 1 r 2 r 2 i u (t)dt = 1 r 2 S + r \S + δr S + r \S + δr u(x, t) 2 G(x, t)dxdt, t u(x, t) 2 G(x, t)dxdt for 0 < δ < 1. The idea at this point is to obtain differentiation formulas for H δ u(r) and I δ u(r), which - by means of a delicate limiting process - will yield corresponding ones for H u (r) and I u (r). In turn, such formulas will allow to establish almostmonotonicity of a suitably defined frequency function. To state this result we need the following notion. Definition 6 We say that a positive function µ(r) is a log-convex function of log r on R + if log µ(e t ) is a convex function of t. In other words µ(e (1 λ)s+λt ) µ(e s ) 1 λ µ(e t ) λ, 0 λ 1. This is equivalent to saying that µ is locally absolutely continuous on R + and rµ (r)/µ(r) is nondecreasing. For instance, µ(r) = r κ is a log-convex function of log r for any κ. The importance of this notion in our context is that Almgren s and Poon s frequency formulas can be regarded as log-convexity statements in log r for the appropriately defined quantities H u (r). Theorem 51 (Monotonicity of the truncated frequency) Let u S f (S + 1 ) with f satisfying the following condition: there is a positive monotone nondecreasing log-convex function µ(r) of log r, and constants σ > 0 and C µ > 0, such that µ(r) C µ r 4 2σ R n f 2 (, r 2 )G(, r 2 ) dx. Then, there exists C > 0, depending only on σ, C µ and n, such that the function Φ u (r) = 1 2 recrσ d dr log max{h u(r), µ(r)} + 2(e Crσ 1) is nondecreasing for r (0, 1). 61

62 Remark 5 On the open set where H u (r) > µ(r) we have Φ u (r) 1 2 rh u(r)/h u (r), which coincides, when f = 0, with 2N u (N u as in (59)). The purpose of the truncation of H u (r) with µ(r) is to control the error terms in computations that appear from the right-hand-side f. Proof. [Proof of Theorem 51] First, we observe that the functions H u (r) and µ(r) are absolutely continuous and therefore so is max{h u (r), µ(r)}. It follows that Φ u is uniquely identified only up to a set of measure zero. The monotonicity of Φ u should be understood in the sense that there exists a monotone increasing function which equals Φ u almost everywhere. Therefore, without loss of generality we may assume that on F = {H u (r) µ(r)} and Φ u (r) = 1 µ (r) 2 recrσ µ(r) + σ 2(eCr 1) Φ u (r) = 1 H u(r) 2 recrσ H u (r) + σ 2(eCr 1) in O = {H u (r) > µ(r)}. Following an idea introduced in [33, 34] we now note that it will be enough to check that Φ u(r) > 0 in O. Indeed, from the assumption on µ, it is clear that Φ u is monotone on F. Next, if (r 0, r 1 ) is a maximal open interval in O, then H u (r 0 ) = µ(r 0 ) and H u (r 1 ) = µ(r 1 ) unless r 1 = 1. Besides, if Φ u is monotone in (r 0, r 1 ), it is easy to see that the limits H u(r 0 +) and H u(r 1 ) will exist and satisfy µ (r 0 +) H u(r 0 +), H u(r 1 ) µ (r 1 ) (unless r 1 = 1) and therefore we will have Φ u (r 0 ) Φ u (r 0 +) Φ u (r 1 ) Φ u (r 1 ), with the latter inequality holding when r 1 < 1. This will imply the monotonicity of Φ u in (0, 1). Therefore, we will concentrate only on the set O = {H u (r) > µ(r)}, where the monotonicity of Φ u (r) is equivalent to that of ( ) r H u(r) H u (r) + 4 e Crσ = 2Φ u (r) + 4. The latter will follow, once we show that d ( ) ( ) r H u(r) C r H u(r) dr H u (r) H u (r) + 4 r 1+σ in O. Now, one can show that r H u(r) H u (r) = 4 I u(r) H u (r) 4 S tufg r + r 2. H u (r) 62

63 The desired result will be obtained by direct differentiation of this expression, and using the aforementioned formulas for the derivatives of H u (r) and I u (r). We omit the details Blow-ups and regularity of solutions Similarly to the elliptic case, the generalized frequency formula in Theorem 51 can be used to study the behavior of the solution u near the origin. The central idea is again to consider some appropriately normalized rescalings of u, indicated with u r (see Definition 7), and then pass to the limit as r 0+ (see Theorem 53). Henceforth, we assume that u S f (S 1 + ), and that µ(r) be such that the conditions of Theorem 51 are satisfied. In particular, we assume that r 4 R n f 2 (, r 2 )G(, r 2 ) dx r2σ µ(r) C µ. Consequently, Theorem 51 implies that the function Φ u (r) = 1 2 recrσ d dr log max{h u(r), µ(r)} + 2(e Crσ 1) is nondecreasing for r (0, 1). Hence, there exists the limit κ := Φ u (0+) = lim r 0+ Φ u(r). (60) It is possible to show that κ is independent of the choice of the cut-off ψ introduced in the extension procedure. Since we assume that rµ (r)/µ(r) is nondecreasing, the limit κ µ := 1 2 lim rµ (r) (61) r 0+ µ(r) also exists. We then have the following basic proposition concerning the values of κ and κ µ. Lemma 52 Let u S f (S 1 + ) and µ satisfy the conditions of Theorem 51. With κ, κ µ as above, we have κ κ µ. Moreover, if κ < κ µ, then there exists r u > 0 such that H u (r) µ(r) for 0 < r r u. In particular, κ = 1 2 lim rh u(r) r 0+ H u (r) = 2 lim I u (r) r 0+ H u (r). We now define the appropriate notion of rescalings that works well with the generalized frequency formula. 63

64 Definition 7 For u S f (S 1 + ) and r > 0 define the rescalings u r (x, t) := u(rx, r2 t) H u (r) 1/2, (x, t) S+ 1/r = Rn + ( 1/r 2, 0]. It is easy to see that the function u r solves the nonhomogeneous Signorini problem with u r t u r = f r (x, t) in S + 1/r, u r 0, xn u r 0, u r xn u r = 0 on S 1/r, f r (x, t) = r2 f(rx, r 2 t) H u (r) 1/2. In other words, u r S fr (S + 1/r ). We next show that, unless we are in the borderline case κ = κ µ, we will be able to study the blowups of u at the origin. The condition κ < κ µ below can be understood, in a sense, that we can detect the growth of u near the origin. Theorem 53 (Existence and homogeneity of blowups) Let u S f (S 1 + ), µ satisfy the conditions of Theorem 51, and Then, we have: κ := Φ u (0+) < κ µ = 1 2 lim r 0+ r µ (r) µ(r). i) For any R > 0, there is r R,u > 0 such that (u 2 r + t u r 2 + t 2 D 2 u r 2 + t 2 ( t u r ) 2 )G C(R), 0 < r < r R,u. S + R ii) There is a sequence r j 0+, and a function u 0 in S + = R n + (, 0], such that ( u rj u t (u rj u 0 ) 2 )G 0. S + R We call any such u 0 a blowup of u at the origin. iii) u 0 is a nonzero global solution of Signorini problem: u 0 t u 0 = 0 in S + u 0 0, xn u 0 0, u 0 xn u 0 = 0 on S, in the sense that it solves the Signorini problem in every Q + R. iv) u 0 is parabolically homogeneous of degree κ: u 0 (λx, λ 2 t) = λ κ u 0 (x, t), (x, t) S +, λ > 0 64

65 In addition to Theorem 51 and Lemma 52, the main ingredients in the proof of Theorem 53 are growth estimates for H ur (ρ) and log-sobolev inequalities. Remark 6 Using growth estimates for H u (r) (where u S f (S 1 + )), it is possible to show that necessarily κ > 1. In addition, we have the following: Proposition 54 Let u 0 be a nonzero κ-parabolically homogeneous solution of the Signorini problem in S + = R n + (, 0] with 1 < κ < 2. Then, κ = 3/2 and u 0 (x, t) = C Re(x e + ix n ) 3/2 + in S + for some tangential direction e B 1. Proof. Extend u 0 by even symmetry in x n to the strip S, i.e., by putting u 0 (x, x n, t) = u 0 (x, x n, t). Take any e B 1, and consider the positive and negative parts of the directional derivative e u 0 v ± e = max{± e u 0, 0}. It can be shown that they satisfy the following conditions ( t )v ± e 0, v ± e 0, v + e v e = 0 in S. Hence, we can apply Caffarelli s monotonicity formula [17] to the pair v e ±, obtaining that the functional φ(r) = 1 r S 4 v e + 2 G ve 2 G, r S r is monotone nondecreasing in r. On the other hand, from the homogeneity of u, it is easy to see that φ(r) = r 4(κ 2) φ(1), r > 0. Since κ < 2, φ(r) can be monotone increasing if and only if φ(1) = 0 and consequently φ(r) = 0 for all r > 0. It follows that one of the functions v e ± is identically zero, which is equivalent to e u 0 being either nonnegative or nonpositive on the entire R n (, 0]. Since this is true for any tangential direction e B 1, it thus follows that u 0 depends only on one tangential direction, and is monotone in that direction. Without loss of generality, we may thus assume that n = 2 and that the coincidence set at t = 1 is an infinite interval Λ 1 = {(x, 0) R 2 u 0 (x, 0, 1) = 0} = (, a] {0} =: Σ a. Repeating the monotonicity formula argument above for the pair of functions max{±w, 0}, where { x2 u 0 (x 1, x 2, t), x 2 0 w(x, t) = x2 u 0 (x 1, x 2, t), x 2 < 0 65

66 we obtain that also w does not change sign. Hence, we get x1 u 0 0, x2 u 0 (x 1, x 2, t) 0 in R 2 + (, 0]. Let g 1 (x) = x1 u 0 (x, 1), and g 2 (x) = x2 u 0 (x 1, x 2, 1) in x 2 0 and g 2 (x) = x2 u 0 (x 1, x 2, 1) for x 2 < 0. Exploiting the fact that g 1 and g 2 are the ground states for the Ornstein-Uhlenbeck operator in R 2 \ Σ a and R 2 \ Σ + a, (with Σ + a := [a, ) {0}) respectively, one reaches the conclusion that κ = 3/2 and that g 1 (x) must be a multiple of Re(x 1 + i x 2 ) 1/2. From this, the desired conclusion easily follows. From Remark 6 and Proposition 54, we immediately obtain the following Theorem 55 Let u S f (S 1 + ) and µ satisfies the conditions of Theorem 51. Assume also κ µ = 1 2 lim r 0+ rµ (r)/µ(r) 3/2. Then More precisely, we must have κ := Φ u (0+) 3/2. either κ = 3/2 or κ 2. We are now ready to state the optimal regularity of solutions of the parabolic Signorini problem with sufficiently smooth obstacles. Theorem 56 Let φ H 2,1 (Q + 1 ), f L (Q + 2,1 1 ). Assume that v W2 (Q + 1 ) be such that v H α,α/2 (Q + 1 Q 1) for some 0 < α < 1, and satisfy v t v = f in Q + 1, (62) v φ 0, xn v 0, (v φ) xn v = 0 on Q 1. (63) Then, v H 3/2,3/4 (Q + 1/2 Q 1/2 ) with v H 3/2,3/4 (Q + 1/2 Q 1/2 ) C n ( v W 1,0 (Q + 1 ) + f L (Q + 1 ) + φ H 2,1 (Q 1 ) ). The proof of Theorem 56 will follow from the interior parabolic estimates and the following growth bound of u away from the free boundary Γ(v) (which, in turn, can be deduced from Theorem 51 by fixing σ = 1/4 and choosing µ(r) = M 2 r 4 2σ ). Lemma 57 Let u S f (S 1 + ) with u L (S + ), f 1 L (S + 1 ) M. Then, H r (u) C n M 2 r 3. 66

67 4.2.4 Regular free boundary points At this point we turn our attention to the study of points on the free boundary having minimal frequency κ = 3/2. Definition 8 Let v S φ (Q + 1 ) with φ Hl,l/2 (Q 1), l 2. We say that (x 0, t 0 ) Γ (v) is a regular free boundary point if it has a minimal homogeneity κ = 3/2. The collection R(v) of regular free boundary points will be called the regular set of v. The following basic property about R(v) is a consequence of the fact that κ does not take any values between 3/2 and 2. Proposition 58 The regular set R(v) is a relatively open subset of Γ(v). In particular, for any (x 0, t 0 ) R(v) there exists δ 0 > 0 such that Γ(v) Q δ 0 (x 0, t 0 ) = R(v) Q δ 0 (x 0, t 0 ). Our goal is to show that, if the thin obstacle φ is sufficiently smooth, then the regular set can be represented locally as a (n 2)-dimensional graph of a parabolically Lipschitz function. Further, such function can be shown to have Hölder continuous spatial derivatives. We begin with the following basic result. Theorem 59 (Lipschitz regularity of R(v)) Let v S φ (Q + 1 ) with φ Hl,l/2 (Q 1), l 3 and that (0, 0) R(v). Then, there exist δ = δ v > 0, and g H 1,1/2 (Q δ ) (i.e., g is a parabolically Lipschitz function), such that possibly after a rotation in R n 1, one has Γ(v) Q δ = R(v) Q δ = {(x, t) Q δ x n 1 = g(x, t)}, Λ(v) Q δ = {(x, t) Q δ x n 1 g(x, t)}, The proof of the space regularity follows the same circle of ideas illustrated, for the elliptic case, in Section 2.8. To show the 1/2-Hölder regularity in t (actually better than that), we will use the fact that the 3/2-homogeneous solutions of the parabolic Signorini problem are t-independent (see Proposition 54). However, in order to carry out the program outlined above, in addition to (i) and (ii) in Theorem 53 above, we will need a stronger convergence of the rescalings u r to the blowups u 0. This will be achieved by assuming a slight increase in the regularity assumptions on the thin obstacle φ, and, consequently, on the regularity of the right-hand side f in (55). Lemma 60 Let u S f (S + 1 ), and suppose that for some l 0 2 f(x, t) M (x, t) l0 2 in S + 1, f(x, t) L (x, t) (l0 3)+ in Q + 1/2, and H u (r) r 2l0, for 0 < r < r 0. 67

68 v = φ Λ(v) R(v) x n 1 = g(x, t) Figure 3: The regular set R(v) in Q δ, given by the graph x n 1 = g(x ) with g H 1,1/2 (Q δ ) and α,α/2 (Q δ ) by Theorems 59 and 61 Then, for the family of rescalings {u r } 0<r<r0 we have the uniform bounds u r H 3/2,3/4 (Q + R Q R ) C u, 0 < r < r R,u. In particular, if the sequence of rescalings u rj converges to u 0 as in Theorem 53, then over a subsequence u rj u 0, u rj u 0 in H α,α/2 (Q + R Q R), for any 0 < α < 1/2 and R > 0. Proof. [Proof of Theorem 59] We only sketch the main ideas of the proof, beginning with the space regularity. For u S f (S 1 + ) and r > 0 define the rescalings u r (x, t) := u(rx, r2 t) H u (r) 1/2, (x, t) S+ 1/r = Rn + ( 1/r 2, 0]. Theorem 53 guarantees the existence of u 0 such that ( u rj u t (u rj u 0 ) 2 )ρ 0. S + R Since κ = 3/2, Proposition 54 ensures that u 0 (x, t) = C Re(x e+ix n ) 3/2 + in S +. For given η > 0, define now the thin cone C (η) = {x = (x, x n 1 ) R n 1 x n 1 η x }. A direct computation shows that for any unit vector e C (η) e u 0 0 in Q + 1, eu 0 δ n,η > 0 in Q + 1 {x n c n }. 68

69 Thanks to Lemma 60, we know that e u rj 0 in Q + 1/2, and thus, undoing the scaling, e u 0 in Q + r η for any unit vector e C (η). The Lipschitz continuity in space follows in a standard fashion. As for the regularity in time, our goal is to show g(x, t) g(x, s) = o( t s 1/2 ). uniformly in Q r η/2. Arguing by contradiction, we assume g(x j, t j ) g(x j, s j ) C t j s j 1/2. Let x j = (x j, g(x j, t j)), y j = (x j, g(x j, s j)) and δ j = max{ g(x j, t j ) g(x j, s j ), t j s j 1/2 }. We consider the rescalings of u at (x j, t j ) by the factor of δ j : w j (x, t) = u(x j + δ j x, t j + δ 2 jt) H u (x j,t j )(δ j ) 1/2. (64) The sequence w j converges to a homogeneous global solution in S, timeindependent, of homogeneity 3/2. Combining this information with the fact that e u 0 in a thin cone, we obtain a contradiction. We next observe that the regularity of the function g can be improved with an application of a boundary Harnack principle. Theorem 61 (Hölder regularity of g) In the conclusion of Theorem 59, one can take δ > 0 so that g H α, α 2 (Q δ ) for some α > 0. The proof of this result relies on two crucial ingredients. The first one is the following non-degeneracy of e u: e u cd(x, t) in Q + δ, where d(x, t) denotes the parabolic distance from coincidence set Λ(v) Q δ. This property, in turn, relies on a suitable parabolic version of Lemma 43. The second ingredient in the proof of Theorem 61 is the following version of the parabolic boundary Harnack principle for domains with thin Lipschitz complements established in [56, Section 7]. To state the result, we will need the following notations. For a given L 1 and r > 0 denote Θ r = {(x, t) R n 2 R x i < r, i = 1,..., n 2, r 2 < t 0}, Θ r = {(x, t) R n 1 R (x, t) Θ r, x n 1 < 4nLr}, Θ r = {(x, t) R n R (x, t) Θ r, x n < r}. 69

70 Lemma 62 (Boundary Harnack principle) Let Λ = {(x, t) Θ 1 x n 1 g(x, t)} for a parabolically Lipschitz function g in Θ 1 with Lipschitz constant L 1 such that g(0, 0) = 0. Let u 1, u 2 be two continuous nonnegative functions in Θ 1 such that for some positive constants c 0, C 0, M, and i = 1, 2, i) 0 u i M in Θ 1 and u i = 0 on Λ, ii) ( t )u i C 0 in Θ 1 \ Λ, iii) u i (x, t) c 0 d(x, t) in Θ 1 \ Λ, where d(x, t) = sup{r Θ r (x, t) Λ = }. Assume additionally that u 1 and u 2 are symmetric in x n. Then, there exists α (0, 1) such that u 1 u 2 H α,α/2 (Θ 1/2 ). Furthermore, α and the bound on the corresponding norm u 1 /u 2 H α,α/2 (Θ 1/2 ) depend only on n, L, c 0, C 0, and M. Remark 7 Lemma 62 is the parabolic version of Theorem 31. Unlike the elliptic case, it cannot be reduced to the other known results in the parabolic setting. We also note that this version of the boundary Harnack is for functions with nonzero right-hand side and therefore the nondegeneracy condition as in iii) is necessary Singular free boundary points The main goal of this section is to establish a structural theorem for the set of the so-called singular points, i.e. the points where the coincidence set {v = φ} has zero H n -density in the thin manifold with respect to the thin parabolic cylinders. This corresponds to free boundary points with frequency κ = 2m, m N. We will show that the blowups at those points are parabolically κ- homogeneous polynomials. As in the approach in [36], described in Sections , the main tools are parabolic versions of monotonicity formulas of Weiss and Monneau type. These are instrumental in proving the uniqueness of the blowups at singular free boundary points (x 0, t 0 ), and consequently obtain a Taylor expansion of the type v(x, t) φ(x, t) = q κ (x x 0, t t 0 ) + o( (x x 0, t t 0 ) κ ), t t 0, where q κ is a polynomial of parabolic degree κ that depends continuously on the singular point (x 0, t 0 ) with frequency κ. We note explicitely that such expansion holds only for t t 0 and may fail for t > t 0 (see Remark 8 below). Nevertheless, this expansion essentially holds when restricted to singular points (x, t), even for t t 0. This is necessary in order to verify the compatibility 70

71 condition in a parabolic version of the Whitney s extension theorem. Using the latter we are then ready to prove a structural theorem for the singular set. It should be mentioned at this moment that one difference between the parabolic case treated in this section and its elliptic counterpart is the presence of new types of singular points, which we call time-like. At such points the blowup may become independent of the space variables x. We show that such singular points are contained in a countable union of graphs of the type t = g(x 1,..., x n 1 ), where g is a C 1 function. The other singular points, which we call space-like, are contained in countable union of d-dimensional C 1,0 manifolds (d < n 1). After a possible rotation of coordinates in R n 1, such manifolds are locally representable as graphs of the type (x d+1,..., x n 1 ) = g(x 1,..., x d, t), with g and xi g, i = 1,..., d, continuous. We now proceed to make these statements more precise. Since the overall strategy is similar to the one described in Sections , we will omit the proofs and focus only on the main differences between the elliptic and parabolic cases. For futher details, we refer the interested reader to [26]. Definition 9 Let v S φ (Q + 1 ) with φ Hl,l/2 (Q 1), l 2. (x 0, t 0 ) Γ (v) is singular if H n (Λ(v) Q lim r(x 0, t 0 )) r 0+ H n (Q = 0. r) We say that We will denote the set of singular points by Σ(u) and call it the singular set. We can further classify singular points according to the homogeneity of their blowup, by defining Σ κ (v) := Σ(v) Γ (l) κ (v), κ l. The following proposition gives a complete characterization of the singular points in terms of the blowups and the generalized frequency. In particular, it establishes that Σ κ (u) = Γ κ (u) for κ = 2m < l, m N. Proposition 63 Let u S f (S 1 + ) with f(x, t) M (x, t) l 2 in S 1 +, f(x, t) L (x, t) l 3 in Q + 1/2, l 3 and 0 Γ(l) κ (u) with κ < l. Then, the following statements are equivalent: (i) 0 Σ κ (u). (ii) any blowup of u at the origin is a nonzero parabolically κ-homogeneous polynomial p κ in S satisfying p κ t p κ = 0, p κ (x, 0, t) 0, p κ (x, x n, t) = p κ (x, x n, t). We denote this class by P κ. 71

72 (iii) κ = 2m, m N. We now state the two main results of this section. Theorem 64 Let u S f (S 1 + ) with f(x, t) M (x, t) l 2 in S 1 +, f(x, t) L (x, t) l 3 in Q + 1/2, l 3, and 0 Σ κ(u) for κ = 2m < l, m N. Then, there exists a nonzero p κ P κ such that u(x, t) = p κ (x, t) + o( (x, t) κ ), t 0. Moreover, if v S φ (Q + 1 ) with φ Hl,l/2 (Q 1), (x 0, t 0 ) Σ κ (v) and we let v (x0,t0) = v(x 0 +, t 0 + ), then the mapping (x 0, t 0 ) p κ (x0,t0) from Σ κ (v) to P κ is continuous. Remark 8 We want to emphasize here that the asymptotic development, as stated in Theorem 64, does not generally hold for t > 0. Indeed, consider the following example. Let u : R n R R be a continuous function such that u(x, t) = t x 2 n/2 for x R n and t 0. In {x n 0, t 0}, u solves the Dirichlet problem u t u = 0, x n > 0, t > 0, u(x, 0) = x 2 n, x n 0, u(x, 0, t) = 0 t 0. In {x n 0, t 0}, we extend the function by even symmetry in x n : u(x, x n, t) = u(x, x n, t). It is easy to see that u solves the parabolic Signorini problem with zero obstacle and zero right-hand side in all of R n R. Moreover, u is homogeneous of degree two and clearly 0 Σ 2 (u). Now, if p(x, t) = t x 2 n/2, then p P 2 and we have the following equalities: u(x, t) = p(x, t), for t 0, u(x, 0, t) = 0, p(x, 0, t) = t for t 0. So for t 0 the difference u(x, t) p(x, t) is not o( (x, t) 2 ), despite being zero for t 0. In order to state the aforementioned structural theorem, we need the following definitions. Definition 10 For a singular point (x 0, t 0 ) Σ κ (v) we define d (x0,t0) κ = dim{ξ R n 1 ξ x α x j t p (x0,t0) κ = 0 for any α = (α 1,..., α n 1 ) and j 0 such that α + 2j = κ 1}, 72

73 Σ 2 2 t = 0 Σ 2 2 Σ 1 4 Σ 1 10 Σ 1 4 t = x 2 1 Figure 4: Structure of the singular set Σ(v) R 2 (, 0] for the solution v with v(x 1, x 2, 0, t) = t(t + x 2 1) 4, t 0 with zero thin obstacle. Note that the points on Σ 1 4 and Σ 1 10 are space-like, and the points on Σ 2 2 are time-like. which we call the spatial dimension of Σ κ (v) at (x 0, t 0 ). Clearly, d κ (x0,t0) integer between 0 and n 1. Then, for any d = 0, 1,..., n 1 define is an Σ d κ(v) := {(x 0, t 0 ) Σ κ (v) d (x0,t0) κ = d}. Definition 11 We say that a (d + 1)-dimensional manifold S R n 1 R, d = 0,..., n 2, is space-like of class C 1,0, if locally, after a rotation of coordinate axes in R n 1 one can represent it as a graph (x d+1,..., x n 1 ) = g(x 1,..., x d, t), where g is of class C 1,0, i.e., g and xi g, i = 1,..., d are continuous. We say that (n 1)-dimensional manifold S R n 1 R is time-like of class C 1 if it can be represented locally as where g is of class C 1. t = g(x 1,..., x n 1 ), Theorem 65 (Structure of the singular set) Let v S φ (Q + 1 ) with φ Hl,l/2 (Q 1), l 3. Then, for any κ = 2m < l, m N, we have Γ κ (v) = Σ κ (v). Moreover, for every d = 0, 1,..., n 2, the set Σ d κ(v) is contained in a countable union of (d + 1)-dimensional space-like C 1,0 manifolds and Σ n 1 κ (v) is contained in a countable union of (n 1)-dimensional time-like C 1 manifolds. For a small illustration, see Fig. 4. The following two monotonicity formulas of Weiss and Monneau type play a crucial role in the proofs of Theorems 64 and